PERRO re 


POOF RAINE A A LAL PLAS PY A IB aGd 


NE SLABS. ne Men te 


FAB fl tom ae 


PLP IMA DD 


Pah Pi npep 


SHORT PRACTICAL RULES ‘ 


FOR 


COMMERCIAL CALCULATIONS, 


INCLUDING 


DIVISION SIMPLIFIED AND ABBREVIATED, 


AND 


SHORT METHODS FOR MULTIPLICATION: ORIGINAL AND INGENIOUS METH- 
ODS IN CANCELLATION, THE RULE OF THREE, PERCENTAGE, 
INTEREST AND DISCOUNT, AND A SIMPLE METHOD 
FOR AVERAGING ACCOUNTS. TO WHICH 


- RXPOSITION OF PROFIT AND LOSS, DIVISION INTO PROPORTIONAL PARTS, 


PARTNERSHIP AND BANKRUPTCY, 


INVOLUTION, COMPOUND INTEREST, ANNUITIES, SINKING FUND, 
AND BOND COMPUTATIONS, STERLING, CHRONOLOGICAL 
CALCULATIONS. A SIMPLE METHOD FOR THE EX- 
TRACTION OF THE CUBE ROOT, 

AND 


4 SIMPLE RULE SHOWING HOW TO DISCHARGE A GIVEN DEBT IN 
SEVERAL EQUAL PAYMENTS, IN A GIVEN TIME, INCLUDING 
PRINCIPAL AND INTEREST, AT A GIVEN RATE PER CENT. 


A 
SIMPLE METHOD FOR ADDITION, 
AND 
PRACTICAL HINTS FOR BUILDERS, ETC. 
BY 


PATRICK MURPHY. 


ALBANY: 
WEED-PARSONS PRINTING COMPANY, 
IgI2 


Entered according to act of Congress in the year nineteen hundred and four, 
By PATRICK MURPHY, 


In the office of the Librarian of Congress at Washington. 


Entered according to act of Congress in the year nineteen hundred and twelve, 
By PATRICK MURPHY, 


In the office of the Librarian of Congress at Washington. 


PLAT IS os 
MATHEMATICS tiopas 
Sm Ind LIOKAL 


BUR 1 AsT On nee 


-= A trivial matter having led the author to make some investiga- 
\ tions in the fundamental rules of Arithmetic, particularly Division, 
—~ and having, in the course of these investigations, discovered some 
novel methods for simplifying and abbreviating Long Division, 
‘» he has concluded to present them in this form to the public, 
flattering himself that they will be found, on perusal, both inter- 

= esting and instructive. 
~ In treating the matter, the author assumes that the reader has 

es a knowledge of arithmetical notation and numeration, the simple 

03 rules, and the principles of Division as given in most Arithmetics 

of the present day, and which will be essential to a proper under- 

standing of the subject; hence, the explanation of technical 
_ terms, definitions, etc., unless when necessary, will be found 

; omitted. 

~\ Division, we are aware, could not be changed without making, 

o at the same time, a corresponding change in Multiplication — one 
being the reverse of the other —hence, some extraordinary re- 
sults will be found in the contractions in Multiplication, to which 
attention is invited. 

The method for Addition will be found advantageous when- 

ever the person engaged in adding is liable to interruptions. 

, The rule for finding the weekly day, which forms the closing 
Dieter of this little work, is quite simple, requiring no arith- 
metical process beyond Division for its solution, and is easily 
remembered, as it requires neither monthly nor centennial ratio 

~ to be added. The rule is original so far as known to the 


——, | AUTHOR. 
; ALBANY, January, 1886. 


a 


* + Oxctnsnrsy 


A 4 


V1 


4 PREFACE, 


NEW EDITION 
REVISED, ENLARGED AND IMPROVED. 


In the preparation of this work, our first attempt at authorship, 
Division Simplified, ete., has formed the groundwork of construc- . 
tion. The present edition contains the principal feature of the 
original work, the Vertical Line, by the use of which, most 
extraordinary results in arithmetical calculations are obtained, 
and problems solved much more readily than by the usual 
methods given in text-books. 

In the new work, we have endeavored to make the use of the 
line more simple and clear, explaining the most. important prin- 
ciples connected with it, and giving numerous examples and 
illustrations. 

Original and Ingenious Methods will be found in Cancellation, 
The Rule of Three, Percentage, Interest and Discount, Averaging 
Accounts, Profit and Loss, Division into Proportionate Parts, 
Partnership, Bankruptcy, and Practical Hints for Builders, which 
are given in the present work. 

In presenting this edition, the Author trusts that, by the 
improvements now introduced, it will be rendered worthy, in a 
much greater degree, of a continuance of the very favorable 
reception which it has already experienced from a kind public. 


Tort AUTHOR TO THE READER. 


Before attempting the simple and novel methods given in Can- 
cellation, the Rule of Three, Interest, Discount, etc., w thorough 
knowledge of our methods for Division and Multiplication will be 
necessary, as without the latter, the simplicity and novelty of the 
former can neither be understood nor appreciated. 

January, 1889. 


PREFACE. 5 


IMPROVED EDITION. | 


Some very important improvements have been made in this 
edition, and fifteen pages of new matter added, which, from long 
experience, the Author knows will be found useful and practical. 
To particularize all that has been done would be tedious and 
unnecessary: suffice it to say that, in the present edition, a more 
comprehensive and lucid explanation of our Simple Method for 
Averaging Accounts is given; Short Rules for converting British 
Sterling to American Currency, and for computing Interest on 
British Sterling; Hints on Interest, and Other Short Methods, 
together with a Simple Rule for finding the face of a note, the 
proceeds being given. This, we will venture to say, is the first 
time the rule. has appeared - in print, and will be found of great 
practical utility. 

The reasons of the rules and operations (a part of arithmetical 
science too generally neglected both in treatises on the subject, 
and in teaching), are fully explained by simple and easy illustra- 
tions and examples; and it is hoped that the subjects will thus 
be rendered intelligible and attractive tothe reader. 

July, 1895. 


ENLARGED NEW EDITION. 


Twenty pages of useful and practical matter, carefully prepared, 
have been added in this edition. 

The article on Interest Simplified cannot fail to interest the 
student. The Short methods for finding Interest on Daily 
Balances, and for changing Commercial into Exact Interest, and 
the reverse; also Short Methods on Tonnage, both Net and Gross; 
and the Short Methods on Trade Discounts, together with numer- 
ous examples and illustrations, and the reasons for those Short 
Methods, will be found worthy the attention of the reader. 

July, 1900. 


@ 


We have added to the present edition a Review on Interest, 
with illustrations and examples, the most complete, perhaps, ever 
given on this subject. We have also made important changes in 
the body of the work, and added fourteen pages of Hints and 
Helps for the Studeat, which will be found both useful and 
interesting. 

August, 1904. 


6 PREFACE. 


REVISED, ENLARGED AND IMPROVED EDITION. 


Several changes of a very important nature have been made in 
the present edition. Twenty-one pages have been thrown out, 
and new matter of a more useful and business character substi- 
tuted. Besides this, Twenty-five new pages have been added, 
consisting chiefly of very simple methods for multiplying together 
numbers of two, three, four, five and more figures, whether whole 
or fractional, with copious examples, illustrations and reasons. 
These methods, so far as known to the Author, are entirely 
original, and will be found exceedingly simple and practical. 

August, 1906. 


Some things of minor importance, contained in former editions, 
have been omitted in this edition, and the space thus gained has 
been filled by the insertion of valuable rules, which cannot fail to 
be interesting to the student. 

July, 1908. 


LATEST EDITION, ENLARGED AND IMPROVED. 


To this, the latest and best edition of the work, we have added 
a Chapter on Involution; Compound Interest, including a Table ~ 
showing the Amount of $1, or £1 sterling from 2% to 10%, for 
any number of years from 1 to 35; and a Simple Rule for 
the Computation of Bonds; and for Sinking Funds; also a Rule 
to Ascertain the Amount necessary to Discharge a Given Debt 
in Several Equal Payments in a given time, including both Prin- 
cipal and Interest, at a given rate per cent. 

In connection with this rule, we have given a Table showing 
the Amount necessary to discharge the Debt of $1, in equal pay- 
ments,- from 2 to 21 years, thereby facilitating the solution of 
such problems. 

Much labor and care have been given to the construction of 
this Table and, so far as known to the Author, it is the first of 
its kind given in an Arithmetical work. 

We have also given a Short Rule for the Computation of Paper; 
and a Simple Rule for the Extraction of the Cube Root. 

The rules are Tersely expressed; the Problems are all solved 
and fully illustrated, so that the Student will have no difficulty 
in mastering the different subjects. 

January, 1912. 


CONTENTS. 


PAGE 
Eee INLINE 5 ov. soir se pan cme stne cece ctcecs panate oa. 
General Principles, Land I] .............ceccee cecereces 9 
PON RG cine worse os 26 es Pen tenn sd We aes fe oladine. s 10 
Principal Wustrated ./..... 25... . cee es eee eee eee er eenenes 11-15 
oR Mee EON SE tt. n giuccie ele e's 49 0b so .ese'e = 15 
Tanne the Decimal of the Remainder. ......,....6.00.00- 17 
Simple Method to find the Decimal .. ...........+-.0.00. 17 
De INGIOlGS elie, fs. le kere c clelo vibe canes cece ences 18 
PUTIOT DION LIUSLTALEC 6c ww eee elt e ow Seveee seme exes 19-24 
Se oe Aaa > ata lals 22-5. «ys o'e dias <0 c <-6 #19 w 0s! 24 
eC O LL lees oe hc pide as eS shalt els pw se 64.0 4 © Son s.9 29 
Problems solved by Rule if I eetayey oe oe pees Ta ye ee 30-35 
Problems solved by Rule IIT .........ccccccocees aitsrel Paes 30-47 
Premarks on Rule Tl)... wc ccc enews nctcaccoesene 38-41 
ORS Pa SS (TEs (2 21 a eT ad perss omits Co 
General Short Method for all Numbers.......... om oscar 49 
Short Method to reduce Square Feet to Acres...........06- a) 
Gross Cost given, to find cost of One Article.............. 57 
Me as a ORL fa ioe oe. ca es Ves s 8s & boar Stce oc 58 
meetnods Of Proof for Division... 66... cas veces eRe Neenia 62-65 
TL IROLE Tate os CWS wos eins ov alsin a ae chain ov bw weeds 65 
To Divide by the Nine Digits in Direct Order ............. 68 
To Divide by the Nine Digits in Reversed Order........... 69 
mepoiviie byiMixed NUMbETS. 0050... wae ee Gk seed eoewe 70 
meoriahcinons tor Multiplication’. 022.6005. pee SO. 71-88 
To Multipy by the Nine Digits in Direct Order........ ss &2 
To Multiply by the Nine Digits in Reversed Order ........ 83 
Methods of Proof for Multiplication ; Figures, 9and11.... ° 88 
RRO MORES TERE Par Soar edi aah kale oho. 5 ae pte wh ao. wis 26 pl lwo 0. wie w/e 89-97 
ities prule.of Three”. o....5. 3 Sct hy OPE ey ere 97-106 
Pee OM TIC  ETOPOTHMON aiy. 4 cece die see ne ss deletes are aveten ters 107-109 
erecvica ) roolems -"lron) Steel ete. ...8 oes esceccsece seca’ 109-113 
ENE PR BTS ito Fs One 4 a xg Wn kb alo CR elewselawiain 6 115-121 
Interest, both Commercial and Exact .... .ssseeesecee eee 122-128 
martial Payments Or INGOrsements .. seine aecc ccs cuceee 129 
ae T ICO iG ale ie ie i fad iets co vin eB nv seats bog e'tens ot 131-136 
Averaging Accounts; Interest Methods...... ........ «- 137-142 
RETRACTS A Sree tees sete 5! aise titi dias «; wade ofoie's v0 5 SOlN Ss 143-148 
Divison into Proportional Parts. .............. eee ener’ ee 149 
Peavitaraisi pin ss). =. bahar gs sete ete CPE nk sade? ema os) aot ove. 6 s4 ® 152 
Pe SAL Ven ah PES alae 25%, Paw AREAL VA WC et «Ne eee ore ae 154 
Deer terctilesee with LilUstrationses <. sickvevescesse eee. abs - 155-158 
Serer Caer IHbe MOT EWP, iereisieicy pei «0 << 64 class 04'e cae 159-173 
Lumber Calculations Simplified...... Se aienes aye pie eflanes aiets 172-173 
MOE ICHU RICUIAUOUS) Oye c rite deeedeesestececds arte 174 
Bissextile or Leap Year Explained ....cocessesscsiecvecces 176 


APPENDIX. 


PAGE 
Interest Rules Tersely Stated and ene 20 ni-eis Sidiole’s ein sleaeretinamnelets 177 
Special Rules — Six per cent. Method.... ... ,.....seeeceres wd ses aia pee ae 180 
Interest on Running ACCounts <2 o. sive. sou - news cele theleseieierete te tne amernae 182 
Important Facts to be Remembered | CaS Rae Sietele eieis, bia) cate Nenejere tele eretaaterene 184 
A Simple Method for Averaging AcCounts.........ccccescsscsceec sce o-o. 185-194 
Short Methods — Steel, Iron, Coal, ete s2..)2... siete 6 sone stein erate ane meets 195-200 
Pounds Reduced to Gross Tons; Short Meee: seach eee a pane epee 200 
Sterling — Pounds, Shillings and’ Pence. oo... se ccs sos van ele renee 201 
Sterling Reduced to American Cee sio.sie oie le via seis aye isles ale eee Seamer 204-206 
interest on’ Sterling’. oe seve soso ce bac Se Soeseare cab wis olerere eee 206-209 
Other Short: Methods hase Ft Seisele tere tee ee roa) oslo 6 sole miaers aeenereeete ee 209-211 
Hints on Interest — Savings Banks, 3/3352 wig 6 bntete'ere stew ¢ ereeeettee Japaeeee eis 
Interest on Monthly Payments............. celestial eileen eer weenie Seer e eee 212 
Interest on Weekly Payments... «Usual » volaldcaleretelete Pusleinte 213 
Simple Method to Find the Face of a Note. .6200e va 2 nes WSR ne eee 214 
Interest for Months at any Rate per cent.......... ...e+eceeee Sie gharsiemiereiens 215 
Interest Simplified ... = .... ... e'sinie eos 0 heels, salto eeneieeleta enema Oe ee) 
Important Facts, Illustrated by Exam ples... sia 8.8 sie.eiaielsoleis ee arate 218 
To Change Commercial Interest to Exact Interest. 3... ee cipoastesie 219 
To Change Exact Interest to Commercial Interest..,... o irelalielata oietee het ete 220 
Interest on Daily Balances — Short Method....: ............ ASIC De 220 
Bank Balances — Subtraction performed by Addition....... songs ate Sanletteiers 221 
Pig-Iron, 2268 lbs. or 2240 lbs. — Short Methods. . oo 0's 616 Ree elton iS tenteren ceemeree 
To Reduce Pounds to Tons of 2268 1bS........-.sss0ccceeceeee. Siew eciersleus ei 224 
To Reduce Pounds to Tons of 2240 Ibs. . bk 's ioe Wiel ete ina /ach anya teh ereter ata eae 225 
Gross Tons Reduced to Net —Short Method..........c.sccceececcececeees 226 
Net Tons Reduced to Gross —Short Method... ........ 00.00 ecen cee vee 227 
The Net Cost being given to find the Gross Cost.........c.cceccccsce cece 223 
The Gross Cost being given to find the Net Cost........ .cccc.cccccees es 228 
Computing the Cost of Pounds by the Net Ton........... Sabice Reale ete veces 229-230 
Gross Tons of Rails to Mile—Short Method..............cceceees BU Goer 231 
Net Tons of Rails to Mile — Short Method...... Stade tateee “sik tonsietarsiniereteieiats 231 
Trade Discounts —Short Methods ~.. 02. 5.2... 2. v2s cane ce cance eireemeemee=caa 
Interest: Reviewed. 2.0.0 90 6 lhe Fe Wess Sele veered WoL smisen 5 elle ee Reet eee 238-245 
When the Rate Changes Fr equentlyoec.win cjoeaieeerieete ose wiesaaatetoen tien 245 
Annual Interest. .-. scare: 6 ca.g Se celele. eralvare rebehotietelateneeseo mt arene 246 
Partial Payments — Merchants’ Rule,.:.:.-..)s-0s-0.. eee alee eeae ets : 247 
Hints:and Helps for the Students... ics ce cocsnl eee eee enemas Saar. 248-280 
Decimal Fractions. © e660 sis ied win woes Je lw tae Se eee 254-257 
To Multiply the ’Teens Together — Short Method.. . .... ..............- 258 
To Multiply the Twenties, Thirties, etc., Short Method............... ... 259 
To Multiply Mixed Numbers. .2:..0:.¢. 0s 83) bcd er 277-279 
A Simple Method for Addition... 5... c0.0cc.0) 2202.2 ccs code sees eee Oe 
A Simple Method for Subtraction ................eeesee- oars sueeteaeme calc renaes 283-284 
Problems with their Solutions . 2.5.52... seeccs 0 cues dele ene ERT tereeearenn 285 
TAV.GLUELON ES poh Les neni eae reat ool nie erent 9 ayereSsin sere piekele of eccliel ele nectar Neer 293-295 
Compound Interest, and Tables... onesie eo cleia) seal levhipie's se evens eTetele tere aaiarerer aoe acct 
Sinking Fund Computations srgieceret 1a'eje\ioia.oca vai clei sicte ee) eievel aia erels emilee epereeate ee eteowes 301 
Bond Computations. 22 16 30s cas oe co Sere cee eee nee Ete eek tegmastirg 302-303 
AMNICIES {Seek ee whi Eee ele sea eae ola cee ee 303 


To Discharge a Given Debt in Several Equal Payments, Including Principal - 
and Interest, at Any Rate Per peace ene Te to ee acy 
Computations. . : ves eee . 804-307 
A Simple Method for Cube Root.. ..s:ss0s+seseces anak -totee oeteene eae 808-309 


DIVISION 
SIMPLIFIED AND ABBREVIATED. 


GENERAL PRINCIPLES. 


There are certain general principles of Division, a 
knowledge of which is essential to a proper understand- 
ing of the simplified methods given in this work. The 


following are some of the most important: 

I, Since the quotient in Division is the result obtained by divid- 
ing the dividend by the divisor, it is evident that the value of the 
quotient depends upon the relative values of the dividend and 
divisor. Hence, | 

Any change in the value of either dividend or divisor must produce 
a change in the value of the quotient ; but if a similar change be made 
in both dividend and divisor, at the same time, the quotient undergoes 
no change. 

IJ. If a number be added to the divisor, the dividend must be 
increased by the product of the quotient and the number so added, 
in order that the quotient may remain the same. Thus, 


84-10 — 8...4 


' Now, if any number, say 2, be added to the divisor, 10, and we 
desire to divide by 12, without changing the quotient of 10, we 
increase 84 by 2 times the quotient 8, or by 16; then, dividing 
the sum, we have 100 + 12 = 8...4, the same as was obtained by 
dividing 84 by 10. 


10 DIVISION SIMPLIFIED AND ABBREVIATED. 


And if the dividend be no¢ thus increased, the quotient will be 
diminished by the result obtained by dividing the product of the 
quotient and the number added by the new divisor. Thus, 


84 +10 —8...4 
Now, if we divide by 12, without increasing 84, we have, 
84 +12—7 


the same as if 2 times 8 were divided by 12, and the result sub- 
tracted from the quotient of 10. Thus, 


84 - 10 — 8...4, less 2 times 8, or 
16 divided by 12: 16+ 12 —T1...4 


7 


Notes. — 1. When the divisor is contained in the dividend without a re- 
mainder, the division is exact. 

2. When the division is not exact, a part of the dividend is left, this is 
called the remainder and must be less than the divisor. 

3. The remainder is always of the same name or kind as the dividend, 
being a part of it. 


The division of one number by another is denoted in severat 
ways. Thus, either of the expressions, 84 + 10 = 8...4; 
S4 — 8...4; or 10)84 means that 84 is divided by 10, that 

8...4 
the quotient is 8, and the remainder 4, fully expressed 8-4). 

The quotient, also, may be expressed in several ways. Thus, 
875 
the last expression, it may be expressed thus, 8 | 4, showing that, 
to divide a number by 10, we simply cut off the unit figure of es 
dividend for remainder. 


8’/,), 8.4, or, if we substitute a vertical line for the point in 


Rute 


To divide by a number expressed by 1, with a cipher 
or ciphers annexed: Cut off from the right of the divi- 


Div1ston SIMPLIFIED AND ABBREVIATED. 11 


dend, by a vertical line, as many figures for remainder, 
as there are ciphers in the divisor. 


Exam. 1. Divide 475891 by 1000. 
475|891 


Here, there are three ciphers in the divisor, and we cut off from 
the right of the dividend, three figures, 891, for remainder, and 
475 is the quotient; fully expressed, 475891, 

Now, let a number, say 7, be added to the divisor, 1000, and let 
the given dividend, 475891, be divided by the sum, or new divi- 
sor, 1007, according to principle II. Thus: 

Dividing first by 1000, we get 475 for quotient, 


and 891 for remainder. 475|891 
To finish the division, now, according to the _ 8/804 
principle referred to, we multiply the quotient, 472|587 
475, by 7, the figure added, and divide the pro- 
duct, 3325, by the new divisor, 1007, as shown in 1007)3325(3 
: : : 3021 
the margin, getting 3 for quotient, and 304 for 304 


remainder. Subtracting this result from 475/891, 
found above, we get the true quotient, 472, and the true re- 
mainder, 587; fully expressed, 472,°8,4. 


The required quotient for 1007, however, may be more 
readily obtained as follows : 


First, divide by 1000; the quotient is 475, 475|891 ~ 1007 
and the remainder, 891. Then, extending the 3/825 
vertical line to a suitable length, multiply the 479/566 
quotient, 475, by the excess, or added figure F1 
7, and set the product, 3325, so that its unit 587 
figure will be under the unit figure of the re- 
mainder, 891, and subtract the said product from 475/891. 

Next, multiply 3, that part of the product (now partial quo- 


1 DIVISION SIMPLIFIED AND ABBREVIATED. ~ 


tient), to the left of the line, by 7, also; set the result, 21, under 
the remainder, 566, and add both; the sum, 587, is the true re- 
mainder, and 472, to the left of the line, is the quotient. 


The reason of the last process will be understood by 
comparing it with that which immediately precedes, in 
connection with the well-known principles: (a) the greater 
the divisor, the dividend remaining unchanged, the less . 
will be the value of the quotient; and (b) the less the 
divisor the greater the quotient. 


The extended vertical line, observe, divides 3325, as well as 
475891, by 1000. Now, the product 3325 being divided by 1000 
in the last process, while it ought to have been divided by 1007, 
as in the first (Prin. ID), the result, 3|325, is too large (Prin. b); 
and subtracting this result, which is too large, from 475|891, leaves 
too little; consequently, we must add the excess taken away. 

This excess, observe, is the difference between 3/325, in the last — 
process, and 3/304, the true result taken away in the first process, 
or 21, which is 7 times 3, the partial quotient, or that part of the 
product, 3/3825, to the left of the vertical line. 

Or the reason may be explained thus: The true result to be 
subtracted is 3/804; we have subtracted 3/3825; the difference is 
21. Having taken away 21 too many we restore the same by ad- 
dition. 


Before giving the rule for this method of division, the 
following four examples are necessary, as they still fur- 
ther illustrate our method and explain some changes 
which will frequently occur when making use of this 
method : ne 


DIVISION SIMPLIFIED AND ABBREVIATED. 13 
Exam. 2. Divide 3487286 by 1006. 


3487|286 
20/922 
3466/364 
126 

490 


In this, we first divide by 1000, as in the previous example, get- 
ting 3487 for quotient and 286 for remainder. Then, multiplying 
3487 by 6, the product, 20922, is set under the first result and 
subtracted; the difference is 3466|364. In the subtraction 1 is 
carried, which makes 21 to the left of the vertical line, and in 
multiplying next by 6, we multiply not 20, but 21 (20 plus the 1 
carried), setting the product, 126, under the remainder, 364, and 
adding; the true remainder is 490, and the true quotient 3466. 


Nore. — The number carried from the remainders to the quotients forms a 
part of the quotient to which it has been carried, and must evidently be 
maltiplied in with it. 


Exam. 3. Divide 348728654 by 1010. 


348728654 
3487/20 
52411374 

34/870 
76/244 
350 
B45275|/894 
10 

904 


Here, we first divide by 1000 as in the other examples, cutting 
off 654 for remainder. Then, multiplying the quotient, 348728, 


14 DivISION SIMPLIFIED AND ABBREVIATED. 


by the excess, 10 (to multiply by 10, we simply annex a cipher 
and copy the figures), the product, 3487280, is set as in the pre- 
ceding examples, and subtracted (omitting the two outside fig- 
ures, 34, for the present, as they undergo no change by the — 
subtraction). 

Next, multiplying that part of the product, 3487, to the left of 
the line, by 10, also, and setting the product, 34870, as before, it 
is added (omitting the figures 52 for the present). : 

In the addition 1 is carried to 34, making 35, which is now 
multiplied by 10, the product 350 set as before and subtracted. 
The figures 3452 are now brought down, giving 345275 for quo- 
tient. 

In subtracting 350 from the result above it, 1 was carried and 

taken from 76, giving 75; and if actually expressed, said 1 would 

be set under 76. It is evident, therefore, that such a figure is a 
partial quotient, the same as 3487 and 34, and must be treated 
accordingly; hence, 1 is multiplied by 10, also, and the product 
10 added to 894, giving 904 for the true remainder. 


Exam. 4. Divide 3509615 by 1012. 


3509/1615 
49/108 
3467/507 
504 
1011 


In this we divide by 1000, as in the other examples, and mui- 
tiply by the excess, 12, getting 8467 for quotient and 1011 for 
remainder. 

In adding the remainders 504 and 507, it would appear that we ~ 
ought to have carried 1 to the quotient’s place, multiplied said. 1 
by 12, as in the preceding example, and subtracted; but we 
observe that the sum, 1011, is /ess than the divisor 1012, and,% 
therefore, 1011 is the final remainder. 


DIVISION SIMPLIFIED AND ABBREVIATED, 15 


Exam. 5. Divide 7423520 by 1040. 


7423/5210 + 1040 
296/92 
126 60 
11|88 
7138|48 
48 


In this, we cut off the cipher from the right of the divisor, 1040, 

~ and also the cipher from the right of the dividend. Then, divid- 

ing 742352, the remaining part of the dividend, by 104, the re- 

maining part of the divisor, as in the preceding examples, we get 
7138 for quotient, and there is no remainder. 

In multiplying 296 and 11 by 4, the excess in 104, it must be 
borne in mind that 1 has been carried from the remainder 92, in 
subtracting, and 1 also from the remainder 88, in adding, and 
that we have multiplied 297 in the one case, and 12 in the other, 
by 4, getting 1188 and 48, 


Nore.—It need scarcely be remarked that the numbers to the right of the 
vertical line are remainders, and those to the left quotients; that is, partial 
remainders and partial quotients, till the final quotient and remainder are 
found. 


Route II. 


From the foregoing examples and illustrations we 
derive the following: 


Rute. To divide by a number which is a little in excess of 100, 
1000, 10000, etc. : 

I. Divide, first, by 100, 1000, 10000, etc., as the case may be, using 
a vertical line to cut off the remainder, as pointed out in Rule I. 
(Call 100, 1000, etc., in such cases, the approximate divisor.) 

Il. Hatend the vertical line to a suitable length. and multiply the 
figures to the left of said line by the excess jigure of the divisor, setting 


16 DIVISION SIMPLIFIED AND ABBREVIATED. 


the product under the first result, so that writs will be under units, 
etc., and subtract. . 

Ill. If a figure or figures of the suid product eatend to the left of 
the vertical line, multiply such figure or figures, also, by the excess 
figure of the divisor, and include in the multiplication, the figure car- 
ried, if any, from the remainder to the quotient, in the subtraction ; 
set the result to the right under the difference already found, and add ; 
and so on, as long as possible, subtracting and adding alternately, and 
including in each multiplication the figure carried from the remainders 
to the quotients in subtracting and adding. 


Note.—When the last product figure, or carried figure, to the left of the 
line, is brought to the right by multiplication, the division is completed, the 
quotient being to the left, and the remainder to the right, of the vertical line. 


The reason why we subtract and add alternately will 
be understood from the following, in connection with the 
principles of Division already laid down. 


By referring to example 3, page 13, it will be seen that, instead 
of dividing 3487280 (the product of the quotient and the figure 
added, or the excess 10) by 1010, the vea/ divisor, we have divided 
by the approximate 1000, thereby getting too large a quotient; 
subtracting that which is too large leaves too little, as has been 
already explained; consequently, something must be added to 
rectify. 

The true result to be added would be that found by dividing 
the product of 3487 and 10, or 34870, by the real divisor, 1010. 
Instead of dividing 34870 by 1010, however, we prefer to divide 
it by the approximate divisor, 1000, getting 84/870, which is 
added. | 

The result thus added being too large (having divided by a 
smaller divisor than the real one), we obtain too much by the 
addition, and, therefore, we must subtract next; and so on, till | 
the parts of all the products to the left of the line are brought to — 
the right by multiplication; whence, the reason of multiplying by 
the excess figure of the divisor, and setting the products as 


DIVISION SIMPLIFL:ED AND ABBREVIATED. 17 


directed by the rule is evident, and the reason for subtracting 
and adding alternately is shown. 
Or the reason might be expressed in general terms as follows: 
Dividing in each case by a quantity which is too small, gives a 
result: in each case too large. Subtracting too large a quantity 
leaves too little — adding too large a quantity gives too much — 
we must subtract and add till the proper correction is made. 


Notre.— Each new correction becomes less than the previous one, till it 
finally disappears. 


To Frnp tur DeEcrMat. 


Should it be desirable to continue the division into decimals, in 
example 8, the process, according to the usual method for Long 
Division, would be to annex a suitable number of ciphers to the 
‘remainder, 904, and keep on dividing by 1010 till the required 
number of decimals is found. Thus, suppose we required three 
decimals in said example: 


1010)904000(. 895 
S080 
9600 
9090 
5100 
5050 


Annexing three ciphers and continuing the division we get .895+ 


SimepLE Meruop or Finping THE DEcIMAL. 
To find the decimal from the remainder by the simplt- 


jied process: Annex, or conceive to be annexed, as 
many ciphers to the remainder as there are decimals re- 
quired, and continue the division as in the first part of 


the process. 
2 


18 Division SIMPLIFIED AND ABBREVIATED. 


Thus, taking the remainder, 904, inexam- 904]... + 1010 


ple 38, again, we conceive three ciphers an- 9/040 
nexed, and continuing the process, asin the gy4i960 
first part of the example, we get .895+ as 100 
before. » BORER. 


Norte.— This will be found more fully explained in a subsequent part of 
the work. 
(GENERAL PRINCIPLES. 


III. If a number be taken from the divisor, the dividend must 
be diminished by the product of the quotient and the number sub- 
tracted, in order that the quotient may not be changed. Thus, 


84 + 12 —7 


Now, if 2 be taken from the divisor, 12, and we desire to divide 
by 10, without changing the quotient of 12, we subtract from 84, 
the dividend, 2 times the quotient 7, or 14; then dividing the 
difference, 70 (84 — 14), by 10 (12 — 2), we have: 

70+10—7 
the same as: 84 + 12 -- 


And if the dividend be not thus diminished, the quotient will 


be increased by the result obtained by dividing the product of the 


quotient and the number subtracted, by the new divisor. Thus, 
84 + 12 — 
Now, if we divide by 10, without diminishing 84, we have: 
84 + 10 = 8...4 


the same as if 2 times 7, or 14, were divided by the new divisor, 
10, and the result added to the quotient of 12. Thus, 


84 -+- 12 — 7, plus 2 times 7, or 
14, divided by 10: 14+ 10 =—1...4 


Seaee 


hae 
ee 


DIVISION SIMPLIFIED AND ARBREVIATED. 19 


From a due consideration of the foregoing principles, 
the following illustrations will be readily understood : 


If 9216487 be divided by 1000, as in example 1, Rule I, the re- 
sult will stand thus: 


9216|487 


Suppose, now, we subtract any number, say 2, from the divisor 
1000, and divide the same dividend, 9216487 (without diminish- 
ing it) by the difference, 998, according to the foregoing general © 
principles. Thus, 


9216|487 
18/468 
9934\955 


Dividing first by 1000, the quotient is 9216, and the remainder 
487. To finish the division, now, according to Prin. III, we mul- 
tiply the quotient 9216 by 2, the figure subtracted from the divi- 
sor, or the difference between 1000 and 998, and divide the pro- 
duct. 18432, by the new divisor, 998, by Long 
Division, as in the margin, getting 18 for quo- 998)18432(18 


tient and 468 for remainder. This quotient 998 © 
and remainder is now added to 9216|487 found ~ 8452 
above, and the division by 998 is completed, T984 
9234 being the true quotient, and 955 the re- 468 
mainder. 


The required quotient for 998, however, can be more 
simply found as follows : 


9216/4857 ~ 998 
18/432 
36 


9234|955 


20 DIVISION SIMPLIFIED AND ABBREVIATED. 


Dividing first by 1000, the quotient is 9216, and the remainder, 
487. Then, extending the vertical line to a suitable length, the 
quotient, or figures 9216, to the left of the line, is multiplied by 
2, the difference between 1000 and 998, and the product, 18482, 
set so that its unit figure will be under the unit figure of the 
remainder, 487, and the other figures in the corresponding places. 
Then, 2 times 18 (to the left of the line), or 36, is set in proper 
position under the remainder, 432; the results are now added the 
same as in simple addition, giving 9234 for the true quotient, and 
955 for remainder; fully expressed, 9234958, 


The reason of this simple process will be understood | 
by comparing it with that which immediately precedes, 
in connection with the principle: the greater the divisor, 
the dividend not being changed, the less the quotient. 


Dividing first by 1000, which is greater than the real divisor, 
998, the quotient obtained is too small and requires an addition. 
The true result to be added, it will be remembered, is the quo- 
tient, 18, and the remainder, 468, obtained by Long Division, by 
dividing 2 times 9216, or 18432 by 998, as shown in the process 
immediately preceding. 

Instead of dividing 18432 by 998, however, as in the said pro- 
cess, we prefer, here, to divide it by 1000. The extended vertical 
line, observe, performs such division by merely setting the pro- 
duct 18/432, as shown in the margin. 

Since 18432 has now been divided by 1000, while it ought to 
have been divided by 998, the result found, or 18|482, is likewise 
too small, and must have an addition also. The result required 
to be added is evidently the difference between the true result, 
18/468, found in the first process, and 18]432, found in the second. 
The difference is 36, or 2 times 18, that part of the product, 
18/432, to the left of the vertical line. | 

The reason of the simplified process might be expressed in gen- 


DIVISION SIMPLIFIED AND ABBREVIATED. 21 


eral terms as follows: Dividing by a quantity which is too large, 
in every case, gives a result which is too small, and we have to 
keep on adding till the proper correction is made. 


To Fino tHe DEcIMAL. 


To find the decimal for the remainder, 955, true, say to four 
places, the usual method would stand thus: 


998)9550(.9569++ 
8982 


5680 
4990 
6900 
5988 
9120 
8982 


By the simplified process, we simply annex, or conceive to be 
annexed, four ciphers and continue the division as in the other 
part of the work, thus: 


9550]... 
19) 100 
38 


.9569| 


Conceiving four ciphers annexed (three of them represented by 
dots), 2 times 9550, or 19100; then 2 times 19 are placed in 
proper position, as in the first part of the process; then, adding 
the results (rejecting that part of the decimal to the right of the 
line, being less than 5), we have .9569, as found by the long 
process. 


Before giving the rule for this interesting method of 
division, the three following examples are necessary to 


22 DIVISION SIMPLIFIED AND ABBREVIATED. 


still further illustrate our method, and explain a few 
simple changes which will frequently occur when making 
use of this method: 


Exam. 6. Divide 4789365 by 998. 


4789|365 
33/523 
231 

7 


4823/126 


Dividing first by 1000, in this, the quotient is 4789, and the 
remainder, 365. Then, multiplying 4789, the partial quotient to 
the left of the line, by 7 (the difference between 1000 and 998), 
the product, 33523, is set in proper position, as has been explained 
in the previous illustration. Next, that part of the product (now 
partial quotient), to the left of the line, or 33, is multiplied by 7, 
also, and the product, 231, set in position. All the figures to the 
left of the line having now been multiplied by 7, the next step is 
to add the several results. Before doing so, we make a short 
mental examination of the partial remainders, or numbers on the 
right of the line, to ascertain whether any thing is to be carried 
to the figures on the left; and we find 1 is to be carried to 83, 
making 34, which ought to have been multiplied by 7, instead of 
33, giving 238 as the true result to be added instead of 231. The 
same result is obtained, observe, by allowing 231 to remain as it 
is, and add 7 times the 1 carried, or 7, placed in proper position. 
The quotient, then, is 4823, and the remainder 126, fully SADE 
48231234. 


DIVISION SIMPLIFIED AND ABBREVIATED, 23 
Exam. 7. Divide 641458207 by 9930. 


64145|820|7 + 993)0 
449/015 
3/143 
eo, 
645979997 
1/9930 
64593|0067 


In this we first cut off the cipher from the right of the divisor, 
and also one figure, 7, from the dividend, which will be the last 
figure of the remainder. ‘Then, dividing the remaining part of 
the dividend by 993, as in the last example, we get 64597 for 
quotient, and 999 for remainder. The 7 cut from the dividend 
is now brought down, making the remainder 9997; this remain- 
der, being greater than the divisor, 9930, contains said divisor 
once more, 1 is added to 64597, giving 64598 for the true quotient 
and 67 is the remainder. 


Exam. 8. Divide 478498963 by 9991. 


47849|8963 + 9991 
43/0641 
387 


47892|9991 


Here, we divide first by 10000, getting 47849 for quotient and 
8963 for remainder. Then, multiplying the figures to the left of 
the vertical line by 9 (the difference between 10000 and 9991) and 
adding, as in the other examples, we have 47892 for quotient and 
9991 for remainder. The remainder, being equal to the divisor, 
contains it once, and there is no remainder, but 1 is added to the 
quotient, giving 47893 for the true quotient. 


24 DIVISION SIMPLIFIED AND ABBREVIATED. 


Roe III. 


From the foregoing examples and illustrations we have 


the following: 

Rue. To divide by a number which is a little less than 100, 1000, 
10000, etc.: | 
I. Dwide jirst by 100, 1000, 10000, etc., as the case may be, using a 
vertical line to separate the quotient from the remainder. (Call 100, 

1000, etc., in such cases the approximate divisor.) : 

Il. Hxtend the vertical line to a suitable length and multiply the 
figures vo the left of the line by the difference between the approximate 
and the true divisor (this difference is called the complement), 
setting the product under the jirst result, so that units will be under 
units, etc. 

Ill. [fa part of the said product extend to the left of the line, mul- 
tiply such part, also, by the complement, setting the product as before, . 
and so on till the figures on the left of the line are exhausted. 

IV. Add the several results; the true quotient will be on the left, 
and the remainder on the right, of the vertical line (not forgetting to 
multiply the figure carried, if any, from the remainders to the quo- 
tients, setting the product in proper positior and adding it, as shown 
in example 6). 


Arithmeticians have given a simple rule for dividing 
by 100, 1000, ete. (Rule I of this work). We have now 
established, in addition to said rule, two very important 
and equally sumple rules, namely: first, for such num- 
bers as are a little in excess of 100, 1000, ete. (Rule IJ), 
and second, for such numbers as are somewhat less than 
100, 1000, ete. (Rule ITI). 

A knowledge of these two simple rules, together with 
the principles of Division, will now enable us to simplify 
Long Division to an extraordinary degree, and, as a mat: 


DIVISION SIMPLIFIED AND ABBREVIATED, 25 


ter of course, problems in other branches of Arithmetic 
where Long Division has to be used. ‘l'o do so, we draw 
very largely on the following simple principle: 

Multiplying or dividing both dividend and divisor by 
the same number, does not change the value of the quo- 
tient. Thus, 

15 + 8 —5, and 4 times 15 divided by 
4 times 3, or 60 + 12 — 5 
Again, 4812 — 4, and the half of 48 divided by 

the half of 12, or 24 6 —4 ; 


PROBLEMS. 


From a due consideration of the foregoing rules, prin- 
ciples and illustrations, the student will readily under- 
stand the following problems: 


Exam. 1. If 26 building lots be sold for $90350, what 
is the average price of each lot ? 

To solve this problem, we simply divide the price by the num- 
ber of lots. 

Now, it is evident that if 4 times as many lots were sold for 4 
times the money, the average price of each lot would still be the 
same; and as we can more readily divide now by 4 times 26, or . 
104 (Rule II), we divide 4 times the money, $361400 by 104, thus: 

90350 + 26 
3614/00 ~ 104 


26 DIVISION SIMPLIFIED AND ABBREVIATED. 


Multiplying both dividend and divisor by 4, we take the pro- 
ducts for a new dividend and new divisor. 

Dividing first by 100, the quotient is 3614; this is next multi- 
plied by 4, the excess, and the product, 14456, subtracted. Then 
145 (144 plus the 1 carried in subtraction) is multiplied by 4, also, 
and the product, 580, added. Finally, 6 (5 plus the 1 carried in 
addition) is multiplied by 4, and the product, 24, subtracted; 
there is no remainder, $3475 being the quotient, or the average 
price of each lot. 


Exam. 2. If 196 tons of iron cost $7 252, what is the 
price of a ton? 


To solve this we divide 7252 by 196, the quotient is the price. 

Now, it is evident that if half 196 tons be bought for half the 
money, the price per ton will be still the same; and as we can 
more readily divide by 98, half of 196, than by 196 itself, we di- 
vide half the money, or $3626, by 98; thus: 


36/26 + 98 
12 
98 
Here we make use of Rule III. Dividing first by 100, the quo- 
tient is 36, and the remainder 26, Multiplying 36, the quotient, 
_ then, by 2 (100 — 98), the complement, and setting the product, 
72, under the remainder, 26, we add. ‘The remainder, here, being 
equal to the divisor, 98, evidently contains said divisor, once 
more, and hence the true quotient is $37, the price per ton (see 
exam. 8). 


Exam. 8. If 3834 suits of clothes cost $12408.10, what 
is the price per suit ? 


Here we sce that the divisor, 384, is nearly one-third of 1000, 


DIVISION SIMPLIFIED AND ABBREVIATED. a0 


and bearing in mind that we have now an easy method for 
dividing by the numbers at either side of 1000, etc., whether a little 
more, or a little less, we divide 3 times the price by 3 times the 
number of suits; thus: 


12408.10 + 334 
37/224 .30 + 1002 
14 


15|030 
30 


Multiplying both dividend and divisor by 3, we get $37224.30, 
and 1002. Dividing first by 1000, we get 87 for quotient, and 
224.30 for remainder; we next multiply the quotient, 37, by 2, 
the excess figure of the divisor, and set the product, 74, under the 
unit figure of the dollars, 224. Then, subtracting 74 leaves a 
remainder of 150, evidently dollars, to .be still divided by 1002. 
Now, besides $150 of a remainder, there is a remainder of 30 
cents, also, or 15030 cents in all, to be divided by 1002. The di- 
vision of the cents is performed the same as the dollars, first 
dividing by 1000, and subtracting 2 times 15, or 30. The result 
is then $37.15, the price per suit. 


Exam. 4. If 47 tons of iron cost $1703.75, what is the 
price per ton ? 

Here, we see at a glance that 2 times 47, or 94, is a simpler 
divisor than 47 itself, so we multiply the cost and the number of 
tons each by two, and operate with the products, thus: 

34/07.50 + 94 
2\04 
12 
36|23(50 
1/38 
a 
24/94 


28 DIVISION SIMPLIFIED AND ABBREVIATED. 


Here we make use of Rule III, dividing the dollars first by 100, 
and adding for the complement (6) we get $36 for quotient, 
and a remainder of $23 to be still divided by 94. Now, there 
is a remainder of 50 cents also, making 2350 cents in all for 
the remainder, to be divided by 94; the division is performed . 
same as on the dollars, first dividing 2350 by 100, and adding for 
the complement (6), and we get 24 cents and 94 remaining; this 
remainder being equal to the divisor, its value is evidently 1, 
making 25 cents. The price is, then, $36.25 per ton. 


Exam. 5. If 202 boxes of cigars cost $875.40, what is 


the cost of 1 box ? 
875.40 + 202 


4/37.70 + 101 
4 


33/70 


Dividing the number of boxes and the cost; that is, the divisor 
and dividend, each by 2, the new divisor becomes 101, by which 
we divide according to Rule II, and we get $4.33, the required 
cost. 3 

In dividing the remainder, $33.70, or 38370 cents, by 101, we 
first divide by 100; the next step is to multiply 33 by 1, set the 
product, 33, under 70 and subtract, but we see by inspection that 
in doing so the next remainder would be less than 5, or less than 
half a cent (the remainder would be .37),-and without proceeding 
farther, we see that 33 is the correct number of cents; the answer 
is, therefore, $4.33. 


Exam. 6. If 721 boxes of cigars cost $3121.93, what is 
the cost of a single box ? 


4 ane eee ar | 
445.99 + 108 
12 
Be 


DIVISION SIMPLIFIED AND ABBREVIATED. 29 


Here, we divide both dividend and divisor by 7, and the new 
divisor becomes 108. ) 

After dividing $445 by 108, there is a remainder of $83.99, or 
3399 cents; to divide 3399 by 103, we would first divide by 100, 
this would give 33 for quotient, and 99 for remainder; the next 
step would be to multiply 33 by 8, and subtract the product, 
99, from the remainder, 99, which would give 0 for remainder; so 
we see without proceeding farther in the example, that $4.33 is 
the correct answer. 


Exam. 7. If 816 hats cost $3468, what is the cost of 1 
hat ? 


3468 = 816 
~4/33.50 + 102 
8 
95. 


In this we divide the terms by 8, and the new divisor becomes 
102. The rest is now plain; the answer is $4.25. 


REMARKS. 


Remarks on Rule I].— The attentive student cannot 
fail to see, at this stage of the work, to what an extraor- 
dinary extent Long Division may now be simplified. 
Confining our remarks here to Rule I, without further 
reference, for the present, to that equally, if not more, 
important Rule III, which will be fully considered here- 
after, we see that: 


If we take the number 800, for instance, every number between 
it and 900, differing by 8, can be simplified when used as a divisor. 

Thus, 808, 816, 824, 832, 840, 848, 856, 864, etc., divided by 8, 
will give 101, 102, 103, 104, 105, 106, 107, 108, etc., for new di- 
visors, to which Rule IT is applicable. 


30 DIVISION SIMPLIFIED AND ABBREVIATED. 


Again, if the numbers between 8000 and 9000, 80000 and 90000, 
etc., be taken, a similar relation will be found to exist. 

Thus, 8008, 8016, 8024, 8032, 8040, 8056, etc., divided by 8, 
wiil give 1001, 1002, 1003, 1004, 1005, 1007, etc., for new divi- 
sors. 

And the same is true with regard to the numbers from 200 to 
300, 300 to 400, 400 to 500, 500 to 600, 600 to 700, ete. 

Thus, 308, 306, 809, 312, 315, 318, etc., divided by 3, would 
give new divisors 101, 102, 108, 104, 105, 106, etc.; and the same 
may be said in regard to 3003, 3006, 3009, 3012, 3015, etc.; 30003, 
30012, 30018. 

And 909, 918, 927, 936, 945, 954, 968, 972, ete., divided by 9, 
give 101, 102, 103, 104, 105, 106, 107, 108, etc., for new divisors, 
to all of which Rule II is applicable; and so with the other series. 
Hence, 

When the last figure or figures of the divisor are a multiple of the 
Jirst figure or figures, the division can be simplified. 


Notre.— When one number is contained in another an exact number of 
times, the less is said to be a measure of the greater; and the greater is 
called a multiple of the less. Taking the divisor, 816, in the last example, 
for instance, 8, the first figure, is contained without remainder in 16, the two 
last figures. 8 isa measure of 16, and 16 is a multiple of 8. 


And when the first figure or figures of the dwisor is 
a multiple of the last figure or figures, Rule II is also 
applicable, as illustrated in the following: 


Exam. 8. If 2001 citizens of New York pay an annual 
_ tax of $704352, what is the average tax to each? 3 


704352 + 2001 
852(176 = 10004 
176 


DIVISION SIMPLIFIED AND ABBREVIATED. 31 


To solve this problem, we divide the whole tax by the number 
of persons, thus: 

Dividing the dividend and divisor, each, by 2, we have 352176 
to be divided by 10004. Dividing by 1000, we get 352 for quo- 
tient, and 176 for remainder. The result thus found being too 
large, we multiply the quotient, 352, by the excess, 4, and divide 
the product by 1000, and subtract the result. This result is found 
by simply taking half of 352, or 176, and setting it in proper position 
under the remainder, 176; the vertical line performing the divi- 
sion by 1000; and the answer is $352, the average tax. Or thus: 


7043520 + 20010 
3852/1760 + 10005 © 
1760 


Annexing a cipher to both dividend and divisor; in other 
words, multiplying each by 10, the new divisor becomes 20010. 
Then, dividing by 2, we have 3521760 to be divided by 10005, ac- 
cording to Rule II, and we get $352, as before. 


Exam. 9. If 804 building lots be sold for $738476.34, 
what is the average price of each lot ? 


1384|\76.34 + 804 
~923)0954-+ + 1004 
4/615 
918/480 
95 
5054 


Here, we divide both dividend and divisor by 8, and the new 
divisor becomes 1004, by which we divide as in example 8. (It 
being immaterial whether we divide first by 8 and next by 100, or 


32 DIVISION SIMPLIFIED AND ABBREVIATED. 


divide first by 100 and then by 8; we have here divided first by 
100, thereby getting the position of the vertical line at once.) 
Multiplying the quotient, 928, by 4, we get 461.5, and dividing 
this by 100 gives 4.615, which is subtracted. In subtracting, 1 is 
carried to 4, to the left of the line, making 5; this is multiplied 
by the excess, 4, also, and the product, 2.5, divided by 100, giv- 
ing .025 as result, which is added. The quotient is then $918.50, 
the average price of each lot. 


Or thus: 738476384 + 804.. 
923|Uy95|4 & 1005]0 
4I615 
918]480 
25 
5054 


Moving the decimal point two places to the right, in both divi- 
dend and divisor (this multiplies by 100), and then dividing each 
by 8, the new divisor becomes 10050, by which we divide, as in 
example 5, page 15, and we get $918.50, as before. 


In applying the foregoing methods, recourse may be had to any 
process whereby the divisor can be reduced toa simple one; always 
bearing in mind that, whatever change is made tn the devisor a similar 
change is to be made in the dividend to preserve the relationship. 

By a simple divisor is meant 10, 100, 1000, 10000, etc.; or any 
number near to these, as 101, 1002, 10009, etc.; 91, 92, 994, 99997, etc. 


Exam. 10. Divide 369940704 by 77784. 


Here, we divide the terms each by 7 and 369940704 + 77784 
multiply the results by 9, getting 100008 for 52848672 11112 
a simple divisor. The quotient is 4756. 756138048 ~- 100008 


38048 


Division SIMPLIFIED AND ABBREVIATED. 33 


Exam. 11. Divide 251076872 by 6668. 


In this, we add one half the divisor and one 251076872 + 6668 
half the dividend to each respectively and 125538436 3334 


the divisor becomes 10002. 8766115308 + 10002 
The quotient is 37654, the remainder, 1$$93, 715322 
being equal to 1, which is added to 37653, and 3765319986 
there is no remainder. 16 
10002 


Nore.— It need scarcely be remarked that the remainder in division must be 
always less than the divisor. In making use of the present methods, however, 
the remainder frequently comes out equal to the divisor; in such cases 1 is 
added to the quotient, as in the example, and there is no remainder, the division 
being exact. 

Sometimes the remainder will come out greater than the divisor; in these 
cases the remainder is divided by the divisor, the result is added to the quotient 
already found and the difference between said greater remainder and the 
divisor, will be the final remainder, as shown in example 7, page 23, to which 
the student is referred. (See example, page 250.) 


Exam. 12. Divide 653134680 by 8888. 
653134680 + 8888 
Here, we add an eighth to each term and 81641835 1111 


obtain 9999 for a simple divisor. . 7347716515 + 
The quotient is 73485, the remainder being de ote 
equal to 1, which is added to the quotient. y 
73484|9999 


NotEe.—If there be a remainder and it is desirable to find the equivalent 
decimal, annex as many ciphers as there are decimal places required and 
continue the division, as in the first part, illustrated in the following: 


Exam. 18. Divide 47032938 by 12501 to five decimal 
places. 


In this, the terms are multiplied by 8and 470382988 + 12501 
the divisor is 100008, by which we divide. 37§9/63504 + {00008 


The quotient is 3762, and the remainder 33408 30096 
to which we annex five ciphers for the number oa 
of decimal places required, and continuing the 3340800000 
division as in the first part, we obtain the deci- 2/67264 


mal .33405, making the quotient 3762.33405, . 83405 |382736 
correct to five decimal places. 


Note 1.-- By annexing ciphers to each successive remainder the division can 
be carried to any desirable length. ‘ 

2. lf there be decimals in the dividend annex them to the remainder instead 
of ciphers, supplying the deficiency by ciphers, if necessary, to correspond with 
the number of decimal places required, as illustrated in the following : 


3 


34 Diviston Smp.iFIED AND ABBREVIATED. 


Exam. 14. Divide 183170503.6482: by 24995 to six 
decimal places. 


Here, the terms are multiplied by 4 
and the divisor becomes 99980, a simple 183170503.6432 + 24995 
one, the complement being 20. Multi- 7326 82014.5728 99980 
plying 7326 by 20, the result is set to 1146520 . 
the right beginning at the units’ place; 40 
next, 20 times 2 (1 plus 1 carried in 7828|285745/72800 


adding) are 40, set in proper position, o7 eee, 
and by addition, we find 7328 for quo- “395509158640 


tient, and 28574 for remainder, to which 
5728 is annexed, also two ciphers to 
make six places to correspond to the number of decimal places re- 
quired. Multiplying 285745, now, by 20 and 57, also, by 20 and 
adding, we obtain .285802, true to six places, leaving 88840 for re- 
mainder. The quotient is 7328.285802+. And by annexing ciphers 
to the remainder, and continuing the division, any required number 
of decimal places may be obtained. 


Nort.—Itis worthy of remark, that any part or any multiple of a simple 
divisor can also be simplified when used as a divisor. Thus, if 44 of 99980= 
3332624 be used as divisor. To divide by 383326% it is multiplied by 3 to get 99980: 
and if 4g of 99980=1249714, or 12497.5 be used, multiplying either number by 8 
gives 99480. 

Again, if one-half of 12501=625014, or 6250 5, be used ar divisor; multiplying by 
2, and the result by 8, will give 100008; if 4 times 12501=50004 be used: annexing 
a cipher to the latter we obtain 500040, and dividing this by 5, gives 100008, a 
simple divisor, and so of other numbers. (For further illustrations, see page 250.) 


Exam. 15. Divide $5296765 among 3001 persons. 


765, the result, 5296|765, is the quotient for ———|__ ; 
1000. Then, dividing this by 3, we obtain $1765)988+ | 
1765(|588+ (the decimal may be continued tf de- ee 
sirable), the quotient for 38000. To rectify the 
result which is too large, 3000 being less than the true divisor, we 
subtract the 1-8000 part: of 1765, or eee and we obtain $1765, 
the required quotient. 

To get the 1-3000 part of 1765, remove the decimal point three 
places to the left and divide by 3. (1.765+3=.588+.) _ 


Here, we cut off the three right hand figures, . 5296/7653 + 8001 
58 


Division SmpiiFIED AND ABBREVIATED. 35 | 


Exam. 16. Divide 21187060 by 12004. 


In this, we divide the terms by 4, and the 21187060 ~ 12004 
divisor becomes 3001, a simple one, by which “52961765 = 3001 
we divide as in the foregoing example, the 76515884 
quotient is 1765. . 588-4 


From the foregoing examples and illustrations it will 
_be seen that the division can be readily simplified when 
the divisors run as follows: 


1202, 1203, 1204, 1206, 12012, etc.; 12002, 12003, 12004, etc.; 
1402, 1407, 14014, etc. ; 1602, 1604, 1608, 16016, etc. 


Also such as 

1212, 1414, 1616, 2424, 3636, 4848, 5656, 6464, etc. 
If it were required to divide, for instance, by 

2408, 24008, 240008, etc., we would divide by 8 and get 301, 
3001, 30001, etc., for new divisors, and the division would be as 
in the last example, and so with the other numbers. 

To divide by 3636, for instance, we would first divide by 6, 
‘getting 606; dividing this in turn by 6 we get 101 for new divi- 
sor, and so with similar combinations, always bearing in mind that 

Whatever operation is performed on the divisor to simplify it, the 
same operation must be performed on the dividend also. 


Prosiems Sotvep sy Rote ILI. 


Exam. 1. If 792 milch-cows be bought for $59400, what 
is the average price of each ? 


59400 + 792 
74125 + 99° 
74 
99 


Bb DIVISION SIMPLIFIED AND ABBREVIATED. 


To solve this problem, we divide the whole cost by the number 
of cows, thus: 

A slight inspection of the divisor, 792, shows that by dividing 
it by 8, we obtain a simple divisor, 99, by which we can readily 
divide according to Rule III. Dividing both dividend and divi- 
sor, then, first by 8, we next divide the new dividend, 7425, 
by the new divisor, 99, getting 74 for quotient and 99 for re- 
mainder. The quotient is, then, $7482, or rather, $75, the re- 
quired price. 


Now, as the mental eye cannot always readily tell 
whether such a number as 792, for instance, is a multiple 
of some particular number, as 8, the following will be 
found a much more rapid way of simplifying such num- 
bers when presented as divisors: 


Let us take 800, and call it the approwimate divisor, in S 
connection with 792, the veal divisor, arranging them as [792 
shown in the margin, and using periods, or dots, instead ate 
of the ciphers in 800. The difference of the divisors is 8. 

Examining the process, now, in the foregoing example, it will 
be seen that we first divided by 8, and then by 100, in other 
words, we divided by 800, using the component factors, 8 and 
100 (8x100=800), getting 74 for quotient, and 25 for remainder. 
This result is too small since we have divided by 800, instead of 
792. To rectify, we have to multiply the quotient, 74, by 8 (the 
difference of the divisors), and divide the product, 592, by 800. 
(Gen. Prin. III, page 18.) 

Now, multiplying 74 by 8, and dividing by 800, is the same as 
multiplying by 1, and dividing by 100 (¢8> being equal to ;4,), 
and that is exactly what was done in the example; we simply 
added +17 of the quotient, 74, that is, .74, or rather |74, the vertical 
line being used instead of the point. The price is, then, $7432, 
or rather $75. 


DIVISION SIMPLIFIED AND ABBREVIATED. 37 


It will be seen upon examination, also, that we have simply di- 
vided the eighth part of the total price by the eighth part of the 
number of cows, that is, we have divided $7425 by 99. To divide 
by 99, we use 100 for approximate divisor, and 1 (100 —99) is 


the complement; so we add 47 of the quotient, .74, or |74. 


Notse.— Complement, in the language of Arithmetic, is what any number 
wants of being a unit of the next higher order. In the above example, 99 
wants 1 of being 100, and we call 1 the complement. If 93, 993, or 9993, 7 
would, in each case, be the complement, etc. 


Exam. 2. If the taxes paid by 6986 persons amount to 
$528351.18, what is the average tax to each ? 


(eee 
528/351.18 + 6986 
CTH Seis. a 14 
150 
62/874 
124 


998 


Taking 7000 for approximate divisor, in this, and arranging the 
divisors, as in example 1, we find their difference to be 14. 

To divide by 7000 now, it is immaterial whether we divide first 
by 7, and then by 1000, or divide by 1000 first and then by 7%, 
the result being the same in either case. We will choose the lat- 
ter course. Cutting off three figures from the right of the dollars 
divides by 1000. Then, extending the vertical line, we divide by 7, 
getting $75 for quotient and $478.74 for remainder, or $75|47874 
as quotient for 7000. The result obtained being too small, having 
made use of too large a divisor, we have to multiply the quotient, 
75, by 14 (the difference of the divisors), and divide the product 
by 7000. Multiplying 75 by 14, and dividing by 7000, is the same 
as multiplying by 2 and dividing by 1000 (744, = 7,255); and as 


38 DIVISION SIMPLIFIED AND ABBREVIATED. 


the vertical line performs the division by 1000, all we have to de 
is, simply to add 2 times 75, or 150, placed in proper position. 

There is now a remainder of $628.74, or 62874 cents, in all, to 
be still divided. Now, as the remainder in Division is always a 
part of the dividend, it is evident that 62874 is part of the new 
dividend, $75|47874; consequently, in dividing 62874 cents, there 
is no need of dividing again by 7 (the original dividend having 
been divided by 7), so we simply cut off three figures, 874, from 
the cents, and multiply 62 (to the left of the line) by 2, as was 
done with $75, and setting the product, 124, to the right, we add. 
The quotient is now $75.62, and a remainder of 998. Since the © 
original dividend has been divided by 7, it is evident that 998 is 
only the seventh part of the true remainder. Multiplying 998 by 
7 gives the true remainder, 6986, which contains the true divisor 
once; 1 is added to 62 cents, making $75.63, the required aver- 
age tax. 

REMARKS. 


Remarks on Rule III. —If the student have followed the thread. 
of our reasoning thus far, he will readily see, on analyzing the fore- 
going example, that we have simply divided the seventh part of the 
dividend by the seventh part of the divisor; that is, we have divided 
$75478.74 by 998 (6986 + 7). To divide by 998 we -use 1000 for 
approximate divisor, and add the complement, or 2 times the 
quotient; 2 times $75 in the first case, and 2 times 62 cents in the 
other. This gives $75.62 and a remainder of 998 for quotient; 
fully expressed, $75.62238, or rather $75.63. 

From a due consideration of the foregoing results we see that: 
When the divisor is such that the difference between it and the next 
higher number ending in ciphers is divisible by the significant part-of 
such higher number, the complement, or multiplying Jigure, will be the 
result obtained by dividing said difference by the said significant part. 

Referring to problem 1, for instance, where we have taken 800 
for approximate divisor, in connection with 792, the true divisor, 
we see that the difference is 8, and that said difference contains 8, 


DIVISION SIMPLIFIED AND ABBREVIATED. 39 


the significant figure of 800, once; 1 is the complement or multi- 
plier. | | 
Again, referring to problem 2, we see that 7, the significant 
figure of 7000, is contained 2 times in 14, the difference of the 
divisors; 2 is then the complement or multiplier. 

For brevity we will henceforth call the multiplier in such cases 
the key. 

The student’s attention is now invited to the following: Tak- 
ing 800, as was taken in the remarks on Rule II, we see that every 
number between that and 700, differing by 8, can be readily sim- 
plified when used as a divisor. 

Thus, 800, 792, 784, 776, 768, 760, 752, 744, 736, etc. 

Again, if we take the numbers between 8000 and 7000, 80000 
and 70000, 800000 and 700000, etc., a similar relation will be 
found to exist. 

Thus, 8000, 7992, 7984, 7976, 7968, 7960, etc.; 80000, 79992, 
79984, 79976, 79968, etc. ; 800000, 799992, 799984, 799976, 799968, 
etc. 

Suppose it were required to divide by 7968, for instance, we 
would take 8000 for approximate divisor, and both divisors would 
stand thus: 


We would now divide by 8000, as in-problem 2, and 4 (82 + 8) 


is the key. 
And if the divisor were 79968, we would use 
8. eK 
79968 


32 
where 80000 is the approximate divisor, and 4 also the key; and 
so of the other numbers. 
And if the numbers between 900 and 800, 700 and 600, 600 and 
500, 500 and 400, 400 and 300, etc.; also between 9000 and 8000, 


40 DIVISION SIMPLIFIED AND ABBREVIATED. 


90000 and 80000, etce.; 8000 and 7000, 80000 and 70000, etc., be 
compared, a similar relation will be found to exist. 

If it were required to divide by 2982, for instance, the divisors 
would stand thus: 


Din 
2982 
18 
3000 is approximate and 6 the key. 


Reviewing the numbers, now, which can be readily 
simplified, when presented as divisors, let us take any 
series, say from 700 to 800, placing those to which Rule 
II is applicable in a column to the left, and those to which 
Rule III is applicable in a column to the right, as below, 
and the key for each in a column between, thus: 


Rule IT, ; Key _ Rule III. 
700 800 
107 1 792 
714 2 (84 
721 3 776 
728 ee 768 
735 5 760 
742 6 752 
749 7 744 
756 8 736 
763 9 728 
770 10 720 
TTT 1a 712 
784. 12 704 
(91 13 696 
798 14 etc. 


To divide by 756, for instance, we would apply Rule IT; first 
dividing both dividend and divisor by 7, we would then divide 


DIVISION SIMPLIFIED AND ABBREVIATED. Al 


the seventh part of the dividend by the seventh part of the divi- 
sor, which, in this case, would be 108. The approximate divisor 
for 108 would be 100, and 8 the key. 

And to divide by 736, for instance, it would stand thus, 


Se2 
756 
64 


800 would be the approximate divisor, and 8 the key (64-+8), and 
so of the other numbers. 


When the difference between the real and the approxi- 
mate divisor is /ess than the significant part of the latter, 
we indicate the division by the fractional form; illustrated 
in the following: 


Exam. 3. It required $563650.88 to pay a certain army, 
giving each man $23.92; how many men were in that 
army ¢ — 

94.. 
563650/88 + 2392 
140912/72 = 8 —= 3 
93485|45334- 


78|28334. 
26 


23563|99|66 
33 


To solve this, we have to find how many times $23.92 is con- 
tained in $563650.88. 


Nore.— The arithmetical student need scarcely be told, that operations on 
decimals are performed as on whole numbers, due attention being paid to 
the proper position of the decimal point. 


42 DIVISION SIMPLIFIED AND ABBREVIATED. 


Moving the decimal point in each two places to the right, in 
other words, calling both cents, throws off the decimals, and we 
have whole numbers at once. i 

To divide by 2392, we take 2400 for approximate divisor; the 
difference is 8, which does not contain 24, the aie part of 
2400, so we indicate such division by the fraction 8, which is 
equal to 4, and this is the key or multiplier. 

We now divide by 2400, first by 100 by means of the line, and 
next by 24. To divide by 24 we use the component factors 4 and 
6 (4x6 = 24), dividing first by 4 and then by 6, and we get 
23485 for quotient and .4533+ for remainder, 3 being repeated 
without end. 

We have now to add 8-2400, or =45 part of the quotient, 23485, 
or, what amounts to the same thing, the ;1, part divided by 3 
The hundredth part of 23485 is 234.85, and the third part of the 
latter is 78.2833+, 3 being repeated without end. Next, the =, 
part of 78, or .26, is added. 

Now, since the vertical line performs the division by 100, right 
through, all we have to do is simply divide the partial quotients, 
23485 and 78, each by 3, setting the results in proper position. 

Adding the several results now, we get 23563 for quotient and 
.9966+ for remainder, 6 being repeated without end. 

To find the correct decimal, we continue the process, simply 
cutting off 66 and ee a third of .99, or .83, placed in proper 
position (that is the 45 part of .99, or .0033, this being the 
value of |33 as it now stands in the work). The remainder is now 
.9999-+, or .9 repeated to infinity, which is equal to1. Adding 1 
to 23563 gives 23564, the number of men, and there is no re- 
mainder. | 


Norte. — To show that .9999, etc., is equal to 1: If we take the digit 9 as 
divisor, and any one of the remaining eight digits as dividend, and express 
the value of such division in the language of decimals, we will, in every 
case, obtain a repetition, without end, of the figure of the dividend. Thus, 
4 — .11111, ete. ; 4 — .7777, etc.; and 


DIVISION SIMPLIFIED AND ABBREVIATED. 43 


Conversely, .11111, etc., = 45 .7777, etc., = 45 and on the same principle, 
.9999, etc., is equal to 1; that is, .9999, etc., = 3, or 1, as in the foregoing 


example. Hence, 

To find the value of a decimal repeated to infinity: Set down the repeated 
figure or figures for numerator, and 9, or as many nines, for denominator, 
as there are figures repeated, and the result is a common fraction equal in 
value to the decimal. Thus, .3333, etc., = 3, or 4; and .9696, etc., = 24, 


etc. 


Exam. 4. Suppose $5083763.16 were to be divided 
among a number of persons, giving each $398.32; how 
many persons would receive that sum $ 


ee 
50837|6316 + 39832 
76254 12709/4079 ~~ 198 —49 
318 53/3778 
2996 
fia bated 
12763|0125 


Here we take 40000 for approximate divisor, The difference of 
the divisors, 168, divided by 4, the significant part of 40000, gives 
42, the key. Cutting off four figures, 6316 (one figure for each 
cipher in 40000), from the right of the dividend, and then divid- 
ing by 4, gives the quotient for 40000, or 127094079. 

We have now to add 42 times 12709. To multiply by 42, we 
use the component factors 6 and 7 (6x 7=42). Setting 6 times 
12709, or 76254, a little to the left, as shown in the margin, we 
multiply the latter in turn by 7, setting the product, 533778, in 
proper position, as the rule directs. Next, we add 42 times 53, or 
2226, placed in proper position; thus: 6 times 53, or 318, is set to 
the left, as in the margin; then 7 times 318, or 2226. 

A short inspection, now, shows that in adding the remainders, 
or numbers to the right of the line, 1 is to be carried to the left, 
or quotient’s place; this 1 is also multiplied by 42, and the pro- 


44 DIVISION SIMPLIFIED AND ABBREVIATED. 


duct set in proper position. Addition now gives 12763 persons, 
and .0125 remaining, which is equal to $5. 


Notr.—To find the value of the remainder, .0125, in such examples as 
the foregoing, it must be carefully borne in mind, that the remainder, in 
Division, is always a part of the dividend; and that whatever operations are 
performed on the dividend and divisor, by way of preparation to simplify 
the division, the reverse of these operations is performed on the remainder, 
to find the true remainder. 

In the foregoing example, the dividend has been divided by 40000, conse- 
quently, .0125 is only the 40000 part of what it ought to be. To restore it to 
its proper value; that is, to the same denomination as the dividend, we 
multiply it by 40000. To multiply .0125 by 40000, we simply move the deci- 
mal point four places to the right, which gives 125, the product for 10000; 
multiplying 125, then, by 4 gives 500, or the product for 40000, the true | 
remainder. 

Now, since the dividend in the example is cents, the true remainder, 500, 
is also cents. The value of .0125 is, therefore, $5. 


When the divisor can be conveniently resolved into component 
factors, we proceed by successive division; illustrated in the 
following: 


Exam. 1. Divide 661740804 by 17946. 


Taking 18000 for approximate divisor, we see that the 18000 
difference between that and the real divisor is 54, which 17946 
contains 18, the significant part of the approximate divisor, 54 
without a remainder; and this being the case, the divisor, 

17946, will also contain 18 without a remainder. 


Now, the factors of 18 are 3 and 6 (®8xX6=18). Divid- 38{17946 
ing 17946 by 8, and the result by 6, we obtain 997, a 6) 5982 


simple divisor. 997 
The factors of 17946, now, are 3, 6 and 997, 3|661740804 

by which we divide in succession as shown 6/220580268 

in the margin. The quotient is 36878, and 3676 ae 

the remainder, 997, which contains 997 once th ah et 

more, 1 is added to the quotient, making 330 

36874, and there is no remainder. (See exam. 369731097, 

8, page 23.) 


Nots.—If 18 be taken from 18000, then from the remainder and from each 
successive remainder, the numbers thus found, viz., 17982, 17964, 17946, 17928, 
17910, 17892, etc., each differing by 18, can be treated similar to the example 
shown, if presented as divisors. (For further illustrations, see examples from 
page 249 to 253.) 


Division SIMPLIFIED AND ABBREVIATED. 45 


Exam. 2. Divide 35372903712 by 240168. 


In this, a slight inspection shows that 4|240168 
168, the right hand figures of the divisor, 6} 60042 
is a multiple of 24, the left hand figures; 10007 


and the factors of 24 are 4.and 6. Divid- 

ing 240168 by 4, and the result by 6, we 4135372903712 

obtain 10007, a simple divisor. The fact- 4 8843225928 

ors, now, of 240168 are 4, 6 and 10007 by ry 

which we divide as shown in the margin, BETDS 10856 set D007 
; ; ; 103) 1709 

The quotient is 147284, the remainder a olnis notes 

10007 being equal to 1 (12997). (See Rule 147283) 9279 


II, page 15, and exam. 2, p. 18.) 728 
10007 


Norr.— It will now be readily seen that, when the divisor is not too large, and 
when the last figures are a multiple of the first figures, said first figures being 
easily factored, the foregoing method can be applied; such numbers as: 24168, 
82192, pape 3200192, 48240, 480240, 72648, 720648, 4590, 45090, 450135, 450180, 450225, 
etc., etc. 


GENERAL SHorT Mrtuop ror ALL Numbers. 


When the foregoing methods cannot be conveniently applied, the 
‘following short method, cutting off 50% of the usual work, can be 
used, and may, perhaps, be preferred in all cases; illustrated in the 
following: 


Exam. 1. Divide 17653762 by 3658. 


Here, we take 17653 for the first partial 3658)17653)762(4826. 
dividend, and drawing a line to the right of 3021/7 


3, we find 4 for the first figure of the quo- 95/36 
tient. We now multiply 3658 by 4; but 22)}202 
instead of writing down the product and ~ 1954 


subtracting, we simply add enough to each 
product, as we proceed, to give the figure of the partial dividend. 
Thus, 4 times 8 are 82 and 1 (setting down 1) are 33 which gives 3 
of the dividend; 3 to carry; 4 times 5 are 20 and 3 are 23; and 2 (re- 
quired to make 25) is set down under 5 of the dividend: 4 times 6 
are 24, and 2 (carried from 25) are 26, set down 0, the figure of the 
dividend being 6; carry 2; 4 times 8 are 12, and 2 are 14, and 8 is set 
down to make 17, The number 8021 is the remainder, to which the 
next figure, 7, is brought down; giving 30217 for next dividend. 


(Continued on page 46.) 


46 Division SIMPLIFIED AND ABBREVIATED. 


Now, 3558 is contained 8 times in 80217: 


Say 8 times 8 are 64, and 3 (set down to 3658)17653|762(4826 
make 67): carry 6; 8 times 5 are 40, and 6 a a 
are 46, and 5 (set down to make 51): carry 92902 
5; 8 times 6 are 48, and 5 are 53, and 9 (set mae 0,7 | 


down to make 62): carry 6; 8 times 3 are 

24, and 6 are 30, nothing set down, the number above being 80; the 
remainder is 953 to which 6, the next figure of the dividend is 
brought down, giving 9536 for partial dividend. The next figure of 
the quotient is 2: say 2 times 8 are 16, set down 0, the figure imme- 
diately above being already 6: carry 1; 2 times 5 are 10, and 1 are 
11, and 2 (set down to make 18): carry 1; 2 times 6 are 12, and 1 
are 12, and 2 (set down to make 15): carry 1; 2 times 3 are 6, and 
1 are 7, and 2(set down to make 9): Proceeding thus, we find the 
next figure of the quotient, 6, and the remainder is 254. 


NotrE.— If it should be necessary to continue the work into decimals, annex 
ciphers as in the usual method, and proceed as before. 


When the divisor ends in ciphers we would proceed 
as in the following : 


Exam. 1. Divide 6569437124 by 998000. 


6569/437|124 + 998|000. 
13]138 
ee 
6582|601124 


Here, we cut off the ciphers from the divisor, and as many fig- 
ures (124) from the right of the ‘dividend for the last figures of 
the remainder. Then, dividing the remaining part of-the divi- 
dend by the remaining part (998) of the divisor, we get 6582 for 
quotient and .601 for remainder. The figures cut from the divi- 


DIVISION SIMPLIFIED AND ABBREVIATED. 4% 


dend are now brought down, and the true remainder is 601124. 
The manner of finding the correct decimals of the remainder 
has been already pointed out. The decimal here found is correct 
to two places (.60). 


When the digits of the divisor are all the same figure 
we would proceed as in the following: 


Exam. 1. Divide 3009581015 by 77777. 


30095|81015 T7777 
429440145 11111 
38694/61305 + 99999 
38694 
99999 


Norr, — It will be readily seen that, when the digits of the divisor are all 
the same figure, a succession of 1’s is obtained by dividing the said divisor 
by one of its digits; and multiplying this quotient of 1’s in turn by 9, gives 
a succession of 9’s by which we can readily divide according to Rule III. 


Dividing 77777, in this example, by 7 gives 11111, and multi- 
plying this by 9 gives 99999, a simple divisor. Performing the 
same operations on the dividend gives 3869461305 for a new divi- 
dend. Dividing this by 99999 gives 38694 for quotient and 99999 
for remainder, which, being equal to the divisor, adds 1 to the 
quotient, making 38695, and there is no remainder. 


Notes. — By referring to the process in the foregoing example, we see that 
the number of figures to be cut from the right of the dividend in such cases 
is equal to the number of digits in the divisor, thus giving the position of 
the vertical line before commencing operations on the dividend. 


Hints ror THE STUDENT. 


From the foregoing example it will be seen that, when 
the divisor rs such that, being divided by a measure, or 


48 DIVISION SIMPLIFIED AND ABBREVIATED. 


exact divisor, we obtain a repetition of any of the nine 
digits, the division can be at once simplified. 


This will be made clear by the following observations: Let us 
take the digits 2, 3, 4, 5, etc., repeated two times, three times, 
etc., or any multiple of such repeated digits, and examine them 
carefully for a few moments, thus: 


22x 6—=—182 222x6—1332 2999 x 6 — 13339 
88x5—=—165 333x5—1665 3333 x 5 — 16665 
44x4—176 444x%4—1776 444059 ey 
Bx T= 885 555x7—3885 5555 x 7 — 38885 
66x38—-198 666x3—1998 6666 x 3 = 19998 © 
[x= 154 TI D2—1584 FIT = Bee 
88x3—264 888x3— 2664 8888x 3 — 26664 
99x3—297 999x3—2997 9999 x 3 — 29997 


Suppose, now, it were required to divide by any of the above 
numbers, say 132, 1332 or 138332; we see by inspection that if 
these be divided by 2, 3, 4, 6 or 12, we obtain a repetition, in 
every case, of a particular digit: 66, 666, 6666; 44, 444, 4444; 33, 
338, 38333; 22, 222, 2222; 11, 111, 1111; any one of which can be 
treated the same as 77777 in the last example. 

And the same is true of the remaining numbers, 165, 176, 385, 
etc. : 

Again, if any multiple of those multiples be used as a divisor, 
such as 


132 x 3 = 396 1332 x 3 — 3996 
165 x 38 = 495 1665 x 38 = 4995 
297 x 3 — 891 2997 x 3 = 8991 


etc., the same treatment is applicable; but for most of such mul- 
tiples more rapid methods have been already pointed out, 


DIVISION SIMPLIFIED AND ABBREVIATED. 49 


To divide by 396, 495, 2997 or 8991, for instance, we would 
make use of the next higher numbers ending in ciphers, as 400, 
500, etc., for approximate divisors, getting the key at once, as has- 
been already explained. (See remarks on Rule III.) 


Note. — Dividing the divisor by one of its digits, when the figures are the 
same, and then multiplying by 9, we need scarcely remark, is the same as” 
multiplying first by 9, and dividing by the digit after. To divide by 1332, 
for instance, we first divide by one of its measures, or an exact divisor, say 
2, getting 666, and this in turn by 6, getting 111, which is multiplied by 9, 
getting 999 for asimple divisor. But dividing by2 and then by 6 is dividing 
by 12(2 X 6=12). Now, by first multiplying 1332 by 9, and dividing after, 
the process will be more readily performed; thus, 1332 x 9 = 11988. 

Taking 12000 for approximate divisor now, in connection with 11988, and 
arranging them as has been already pointed out, they willstand asin j9 . 
the margin, the key being 1, andthe division being the same as above; 11988 
that is, by 2, 6 and 1000; or, by 1000 first, and then by 12, as inthe 12 
last arrangement. In such cases, then, it is perhaps preferable to multiply 
first by 9. 


When the digits of the divisor are all the same figure 
and a cipher intervenes, as 10101, 20202, 3030308, ete., 
we would proceed as in the following: 


Exam. 1. Divide 8045261828 by 80808. 


8045261828.. + 80808.. 
301480/920972 + 7999992 
37685/115121+ 
: 37685 
152806 


Here, we see at a glance, that, if the divisor, 80808, be multi- 
plied by 11, we get 888888, by which we can readily divide as in 
the previous example, first dividing by 8, and then multiplying 
by 9. 
Now, multiplying first by 11, and afterward by 9, is multiply- 
4 


50 DIVISION SIMPLIFIED AND ABBREVIATED. 


ing by 99. So we prefer to multiply 80808 by 99 first, vieetting 
. 7999992 for a new divisor. 

To multiply by 99, we conceive two ciphers, represented by 
dots, or periods, annexed to the divisor, as seen in the margin . 
this multiplies it by 100, giving 80808.. as the result. From this 
we subtract 80808, without setting down the latter, however, but 
simply setting down the difference, 7999992, thus: 8 from 10 _ 
(represented by the first dot to the right) and 2; carry 1 to 0 (to 
the left of right hand 8), 1 from 10 (the second dot), and 9; carry 
1 to 8 (the second or middle 8), 9 from 18, and 9; and so on, all 
from the expression 80808. . | 

Going through a similar process with the dividend, that is, 
subtracting 30452618 from 30452618.., without setting down the 
first (being already contained in the last expression), we get a new 
dividend which is divided by 7999992; 8000000 being the approxi- 
mate, and 1 the key, giving 37685.152806 for quotient, true to 
six places of decimals (bearing in mind that the lines is used for 
the decimal point). 

Note.— When two or more ciphers intervene, as 1001001, 7007007, etc., 
we multiply by 999, 9999, etc. (short method), always one 9 more than the’ 
number of intervening ciphers, to simplify the division. (For short method 
of multiplying by any number of o 8, see Contractions in Multiplication, 
page 74.) 


When the figures of the divisor are repeated; thus, 
212121, 323232, etc., we would proceed as in the last’ 
ease, as illustrated in the following: 


Exam. 2. Divide 16299801882 by 212121. : 


16299801882... + 212191.. 
8)1613680|386318 + 20999979 
7)537893|462106 


76841/923158 
76841 


999999 


DIVISION SIMPLIFIED AND ABBREVIATED. 51 


A moment’s glance at the divisor, in this example, shows that, 
if it be divided by 3 or 7, we get 70707, or 30303, and the divi- 
sion cap be performed as in the previous example. 

Multiplying both dividend and divisor, then, by 99, the new 
divisor becomes 20999979, with which we use 21000000 as ap- 
proximate divisor, the difference being 21 and the key 1. 

The quotient is 76842, there being no remainder. 


Nort. — There aretwo ways for finding the value of the remainder .999999. 

First. Reverse the process performed on the dividend; that is, divide by 
99 and multiply by 21 (3 x 7). Dividing 999999 by 99, we get 10101, and 
multiplying this by 21 gives 212121, which contains the divisor once; 
.999999, therefore, is 1. 

Second. Annex ciphers and continue the division, as has been already 
pointed out, and 9 will be repeated to infinity; but it has been already 
shown that .9 repeated to infinity is equal to 1. 

Whenever, therefore, .9, .99, etc., is remainder in such cases, add 1 to the 
quotient and there is no remainder. 


Orner StmpLtE Meruops. 


It must be carefully borne in mind that the two prin- 
cipal rules laid down in the beginning of this work are 
the entire secret to our methods, a little practice enabling 
us to bring any ordinary number used as a divisor within 
one or other of those rules. Hence, the student need not 
expect to understand what follows unless he has thor- 
oughly posted himself on the groundwork. 

To still further assist the student to master Simplified 
Division, we subjoin a few more examples illustrating 
our methods, which, if rightly understood, will prove ex- 
tremely simple and interesting. 


Exam. 1. If 73 musical instruments cost $12802. 74, 
- what did one instrument cost at that rate ? 


52 DIVISION SIMPLIFIED AND ABBREVIATED. 


From several methods of simplifying the division by 
73, we select the following, as, perhaps, the most simple 
and expeditious that can be given. 


1280274 ae 
426758 2433-+-+ 
42675 .8 243-+- 
4267.58 24-++ 
175|3975.38 + 10001 
175 
38]0088 
| 38 
Annexing two ciphers (represented by dots) to 78, in other 
words, multiplying it by 100, we get 7300. To multiply the divi- 
dend by 100, we simply call the dollars cents, and we have 
1280274 cents to be divided by 7300. To divide by the latter, we 
set under it, one-third of itself, one-tenth of that third, and one- 
tenth of that tenth, and adding, we get 10001 for a simple divi- 
sor. Going through a similar operation with the dividend. we 
get 1753975.38 for new dividend, which being divided by 10001, 
gives 175 for quotient, and .3800 for remainder, to which decimal 
88 is annexed, and continuing the process, we find 38 is the cor- 
rect decimal, or cents, in this case. The price, then, is $175.38, 


Notse.— If the division of 7300 be continued, we find the 7300 
decimal to be .383, etc., as shown in the margin, 3 being nee 
repeated without end. And in adding a tenth, ete., 8 will be 24333 4 
repeated in like manner, so that when the results are added, 9000.999 
we get 10000.999, etc. Butit has been shown (note, page 42) 


that .999, etc., is equal to 1; therefore, 10000.999, etc., = 10001. Hence, 


To divide by 73 then: Multiply the dividend by 100 (simply con- 
ceive two ciphers annexed), and under the product write one-third of 
itself, one-tenth of that third, and one-tenth of that tenth, and divide 
the sum by 10001. 


DIVISION SIMPLIFIED AND ABBREVIATED. 53 


Exam. 2. The annual income of a certain American 
citizen is $1538475 ; what is his income for one day ? 


1538475 + 365 
3076950. + 730. 
10256500 
1025650 
102565 


4215|4215 
4915 


To divide by 365 we double it, then double the dividend, and 
we have 3076950 to be divided by 730. Annexing a cipher (dot) 
now, to both dividend and divisor, we get 7300 for new divisor, 
by which we divide as in the previous example, and we get $4215, 
the income for one day. 


Exam. 3. If the aggregate annual tax paid by 73000 
tax payers be $31220640, what is the average individual 
tax ¢ 

312/20640 + 73000 
104/06880 24333 
10/40688 2433 


1/04068 943 
497|\72276 + 100010 
4270 
68 


Here, we set under both dividend and divisor one-third of each, 
one-tenth of that third, and one-tenth of that tenth, and adding, 
we have 42772276-++, to be divided by 100010. The quotient is 
found by Rule II to be $427.68, the average tax, 


Norr. — We see by inspection that 68 is the correct decimal, or number 
of cents; for if we complete the subtraction and continue the process we 


54 DIVISION SIMPLIFIED AND ABBREVIATED. 


get .68006, to which we annex 80, the figures omitted in the division by 10, 
and we have then .6800680 to be divided by 100010, and the result is .68, as 
found in the example. 


Lo divide by 73000, then, we have the following simple rule: 
Under the dividend write one-third of itself, one-tenth of that third, 
and one-tenth of that tenth, add the four lines together and divide the 
sum by 100010. 


Norz. — By continuing the decimals, in the division of 73000 by 8 and 10, 
it will be seen that 3 is repeated to infinity, and the sum of the four lines 1s 
100009.9999, etc.; but this has been shown to be equal to 100010. (See note 
to example 3, page 42.) It is optional whether the vertical line be drawn 
before taking the parts or when commencing to divide by 100010, as, know- 
ing the divisor, we know the number of figures to be cut from the right of 
the dividend, namely: five. 

This simple and expeditious method of dividing by 73000 will be of great 
value in solving problems in interest, for days, at any rate per cent, on the 
basis of 365 days to the year, (Explained in the article on Interest.) 


Exam. 4. If 732 building lots be valued at $356484, 
what is the average price of each lot? 


356484. + 732. 
1188280. 2440 


118828 244 
487|1948 + 10004 
1948 


Here, we simply conceive a cipher annexed to both dividend 
and divisor, and under each result set one-third of itself, then 
one-tenth of that third, and adding we have 4871948 to be di- 
vided by 10004. The quotient by Rule II is $487, the average 
price of each lot. 


Division SIMPLIFIED AND ABBREVIATED. 55 


A knowledge of the methods here given, will be of particular 
advantage in operations where the divisor is not subject to change, 
as,forexample 43560 sq. ft. to an acre; 5280 fect to a mile; 625 sq. 1. 
in a pole; 7.92 in. to a link; 243, or 24.75 cub. ft. to a perch of 
masonry; 2240 lbs. to a gross ton; 144 articles to a gross, &c. 

In problems on Percentage the methods will be found extremely 
valuable where the divisors run like 91, 92, 93, 94, 95, 96, 97, 98, 99, 
101, 102, 108, 104, Xe. 


Exam. In 76824763 sq. ft. how many acres, true to five 
places of decimals ? 


Norr.— Here, we have to divide by 48560, and 4.0 |4856.0 
a slight inspection shows that the component factors 11/1089 — 
are 40, 11 and 99 (40 x 11 x 99 = 43560.) 298-6 
Pointing off one figure and dividing 7682476.3 
by 4 divides by 40; next, we divide by 11, 1920619.075 
getting 174601.734. This is now 1746)/01.734 
divided by 99 by simply pointing off _ 17/46 
two places for 100, and adding for cen 17 
the complement 1, once 1746 and 1763|647|34 
once 17 set in proper position, the 647 
result is 1763|/64734 6 
65387 


To find the complete decimal, now, 
we divide 64734 by 99; simply adding once 647 and once 6; the 
answer is 1763.65387 acres true to five places of decimals. 


Notet.—To divide by 5280; the component factors are 60, 8 and 11; make use of 
successive division, decimally. To divide by 625, multiply by 4 and the result by 
4, and divide by 10.000 (625 x 4 = 2500, and 2500 x 4= 10000.) bearing in mind to 
multiply the dividend also, to preserve the relation. For 792 take 14, of the 
dividend and divide by 99 (792+ 8 = 99.) To divide by 2434, or 24.75, multiply 
by 4 and divide by 99 (2434, or 24.75 x 4=99.) The component factors of 2240, 
are 40, 7 and 8 (40 x 7 x 8 = 2240.) and for 144 multiply by 7 and divide by 1008 
(144 x 7= 1008.) (See rule and exam. page 225.) 


i 
56 Division SIMPLIFIED AND ABBREVIATED. i 


In many cases where, at first glance, it may appear difficult to 


simplify the division, a short inspection of the divisor will saggest 
amethod, ‘Take, for instance, the following 


Exam. Divide 49862538 by 7854, true to seven places 
of decimals. 


Here, a short inspection 49862538 > 7854 


4532958 714 
of the divisor shows that 9065916 -1428 
it is divisible by 11, giving 6346/1412 + 9996 
2/5384 
714; doubling this gives Seal eee 
6348/6804000/0000 
1428; the sum of the three 2721/6000 
1|0884 
numbers gives 9996 for a 4 
.6806722/6888 


simple divisor. Dividing 
the dividend by 11, in like manner, doubling the result and adding, 
we have 63461412 to be divided by 9996. Dividing now by 10000, 
and adding for the complement 4 times 6346 and 4 times 2 we get 
6348/6804. _ 

To get seven places of decimals, now, we annex seven ciphers 
(one for each decimal required) and dividing 68040000000 by 9996, as 
in the first part of the example, we get 6348.6806722, as required. 


Notrre.—It need scarcely be observed that the division may be carried to 
any length by simply annexing as many ciphers as there are decimals 
required, dividing in each case by 9996, the new divisor. 

Suppose it were required to find five more decimal places; annex five 
ciphers and divide 688800000 by 9996. 


(See examples on pp. 58 and 59.) 


— oe 


Division SIMPLIFIED AND ABBREVIATED. 5% 


The following simple rule will be found useful when the cost per 
gross is given to find the cost of a single article: 


RuLE.— Multipiy the cost per gross (up to $142) by 7; point off three 
figures for decimals, and the result is the cost of a single article, near 
enough for practical purposes. 


Exam. If a gross of padlocks cost $23.50; what is the 
cost of 1? Answer 16c. 


Thus: 23.50 x 7 = 164.50 and pointing off three places we have 
.16450, or 16 c. for business. 


Exam. What is the cost of 1 article at $19 per gross ? 
Answer 13c. 


Thus: 19 X 7 = 1838, and pointing off three places we have .138, 
or 13c. for business. 


Reason.—To divide by 144 we multiply by 7 to get 1008 for a 
simple divisor; and multiplying 19, also by 7, to equalize, we have 
133 + 1008; and we simply divide by 1000 instead of 1008 which 
gives the result near enough when the cost does not exceed $142 
per gross. 


Notr.—If greater accuracy be required set 8 times 133 three places to the 
right and deduct, thus: 133. . 
1064 


and, the result is correct to four places, viz. .131936 
1319 


And if it should be required to find the result correct to six places, we annex 
six ciphers, or one cipher for each decimal required, and divide by 1008. 


Exam. Divide $19 by 144, true to 6 places of deci- 
mals. 


Multiplying both terms by 7 we have 1338 =+- 1008 true to 6 places 
of decimals, thus: 


Annexing 6 ciphers we apply 133000,000 ~- 1008 
1064;000 
rule 2 (page 15) and we get 131986,000 
__—*851e 


1381944, true to 6 places -181944/512 


DECIMALS. 


It has been already remarked that operations on deci- 
mals are performed as on whole numbers, due attention 
being given to the point, or characteristic of the decimal. 

To illustrate, we will take a few of the most Boge 
numbers used in Practical Mathematics. 


Exam. 1. Divide 4985.7192 by .7854. 


49857192 +. 7854 
7122456 1129 
6474/96 + 102 

129|48 
6345/48 
2158 
6348106 
06 


Moving the decimal point in both dividend and divisor four 
places to the right; in other words, multiplying each by 10000, we 
have whole numbers. A moment’s inspection of the divisor, now, 
shows that it is divisible by 7, giving 1122, which, in turn, is 
divisible by 11, giving 102 for a simple divisor. 7 

Dividing the dividend, now, by 7, and the result by 11, we 
have 647496 to be divided by 102. Here Rule II is applicable, 
and we get 6348 for quotient. 


DECIMALS. 59 


Or thus: 
49857192 7854 
71229456 1122 
6474960 1020 
6345/4608 - 9996 
2153880 
8 


6347/9996 


Under the divisor set one-seventh of itself, or 1122; then con- 
ceiving a cipher annexed to this, we have 11220; dividing this by 
11, we get 1020, which is set under 1122, and adding the three 
numbers together, we get 9996 for a simple divisor. Going 
through a similar process with the dividend, we get 63454608 to 
be divided by 9996. Here, Rule III is applicable, and we get 
6348 for quotient, the remainder, 9996, being equal to 1. Hence, 
the 

Rue. To divide by .7854: Below the dividend set one-seventh of 
itself, and one-eleventh of that sevent., setting the latter one place 
farther to the left than its proper position ; add the three numbers 
together, and divide the sum by 9996. 


Exam. 2. Divide 4985.7192 by 3.1416. 


49857192 + 31416 
12464298 + [xd4 
1780614 
1618740 
1586/3652 + 9996 
6344 
‘19996 


Dividing both dividend and divisor in this by 4, we get .7854 
for a new divisor, by which we divide according to the foregoing 
rule, and we get 1587 for quotient. 


60 DECIMALS. 


Exam. 3. Divide 496.258 by .07958, true to four 
places of decimals. 


Here, we move the decimal point in 49695800 + 7958 
the divisor five places to the right, and we “69031995 + 9943 


have 07958., or rather 7958, for a new di- 321565 .75 
visor. Then, as there are only three places 168 

vf decimals in the dividend, we annex two 6935/9587[5... 
ciphers to fill the deficiency, and, moving 501331. 75 


the point five places to the right, we have 6935 9638 
49625800 for a new dividend. 

Norte.—If 8000 be taken as approximate for 7958, andarranged thus: 8... 
514 (42 + 8) isthe key. From this we conclude that if 7958 be divided 7958 
by 8, a divisor will be obtained lacking only 51/ of being some power 
of 10 (in this case 1000). 

‘Dividing both dividend and divisor, then, by 8, the new divisor 
becomes 994%, the complement being 54, as shown in note. Here, 
Ruie III is applicable, 54 times 6203, or 32/565.75; then 5} times 
32, or |168, being set in proper position and added, giving 6235 
for quotient and |958.75 for remainder. 

To find the decimal, now, true to four places, the vertical line 

is drawn to the right of the fourth figure, 7, of the remainder, 
and two ciphers (dots) annexed to the remaining part (so as to 
‘give three places, |[5.., to the right of the line, as in the first 
part of the example); then, setting 5} times 9587, or 50|331.75, 
in position, as in the first part, and adding, we get 6235.9638, for 
quotient, true to four places of decimals (1 being allowed for the 
part of the decimal cut away). 

“Exam. 4. A perch of masonry contains 242 or 24.75 
cubic feet. How many perches of masonry in 86006.25 
cubic feet ¢ | 

86006 .25 + 24.75 

3440|25|00 + 99/00 
34/40 
34 


3474/99 


DECIMALS. 61 


Multiplying both dividend and divisor by 4, in this, we have 
344025 to be divided by 99. Here, Rule III is applicable, 100 
being the approximate divisor and 1 the key. Dividing by 100, 
and adding once each partial quotient; that is, the 1-100 part, 
we get 347422, or rather 3475, the required number of perches. 


Notr. —If we used the fractional form, 2484, we would multiply by 4 
also, 4 times 2434 being 99. Hence, the 


Rute. To divide by 24%, or 24.75; Multiply the dividend by 4, 
and divide the product by 99. 


Exam. 5. Divide 478932673 by 16667 to five places of 


decimals, 
478932673 + 16667 
28735|960388 ~ 100002 
57470 
38564). . 


28735.38567 


Here, we simply multiply both dividend and divisor by 6 (the 
divisor being nearly one-sixth of 100000), and the new divisor is 
100002, by which we divide, getting 28735 for quotient and 
38568 for remainder. To find five places of decimals, now, a 
vertical line is drawn to the right of the fifth figure of the remain- 
der, and as many ciphers conceived to be annexed as there are 
figures cut from the right of the new dividend, namely, five; 
then, continuing the multiplication by 2, as before, we get 
.388567, the number of decimals required. The quotient, then, is 
28735 . 38567. 

Upon examination, however, it will be found that the decimal 
in the present example is not only correct to five places, but it is 
correct to nine places, viz.: .385672286, and the correction may 
be made to any required number of decimal places by continuing 
the process as above. Hence, 


62. METHODS OF PROOF. 


To find any particular number of decimal places we have the 
following simple 


Rue. To the right of as many figures of the remainder as there 
are decimals required draw a vertical line, and to the figures on the 
right of said line, if any, annex, or conceive to be annexed, as many 
ciphers as will make the number of places (both figures and ciphers) 
equal to the number of figures cut from the right of the dividend at 
Jirst, and continue the process as in the first part. 


Nore.— Should the remainder not contain a sufficient number of figures 
for the number of decimals required, annex ciphers to fill the deficiency, 
and proceed according to the rule. 

If, for instance, seven places of decimals were required in the last example, 
instead of five, the line would be drawn two places farther to the right (fill- 
ing up with ciphers, or dots to represent them); then, annexing two more 
dots to the right, so as to make the correct number (five) to the right of the 


fine, proceed as before. 


METHODS OF PROOF. 
There are two principal methods of proving Division: 


First, by multiplication: Multiply the quotient by the 
divisor, and to the product add the remainder, if any; 
the result, if the work is correct, will be equal to the 


dividend. 


Second, by casting out the 9’s. This method of proof, 
which is very easy and convenient in practice, and gen- 
erally preferred by the experienced arithmetician, is given 
in connection with the following: : | 


METHODS OF PROOF, 63 


Exam. How many times is 83 contained in 2869086 ? 


2 2869086 + 83 3 
84429/032 + 996 
137/716 
548 
4 , 
" 34567|300 + 12 — 95 rem. 3 


To divide by 83, we multiply both dividend and divisor by 
12, and the new divisor is 996, to which Rule III is applicable. 
The quotient is 34567, and the remainder, 300, which is divided 
by 12, to get the true remainder, 25. 

Proor. Commencing with the divisor, 83, we add its digits 
from right to left, rejecting 9 from the sum; thus: 3 and 8 are 
11; 9 from 11 and 2; the remainder, cr excess, 2, is now carried 
to the left and reserved. 

The digits of the quotient are next added, and 9 rejected, in 
like manner; thus: 7 and 6 are 13, 9 from 13 and 4; then this 4 
and 5 are 9, and taking 9 from this, leaves 0; next, 4 and 3 are 7, 
from which 9 cannot be taken; 7 is carried to the left, as shown 
in tlie margin. 

Multiplying the excesses, 7 and 2, now, gives 14; the digits of 
this are added, in like manner, making 5, which, being less than 
9, is added to the digits of the remainder, 25, from left to right; 
thus, 5 and 2 are 7, and 5 are 12; then, 9 from 12 and 3, the excess, 
is carried to the right and reserved. 

Finally, adding the digits of the dividend, from left to right, 
also; omitting 9 whenever it occurs, and rejecting 9 from the 
sums, as often as they make 9, or more, we get an excess of 2, 
which is equal to the excess found from the remainder, and the 
work is supposed to be correct. 

Rute. (1) Cast the 9’s from the divisor and quotient ; set the ex- 
eesses to the left of the work and reserve them. (2) Multiply said 
excesses, and cast the 9’s from the product ; add the excess here found, 


64 Merruops oF PRoor. 


to the remainder, and cast the 9’s from the sum, reserving the excess 
to the right of the work. (8) Cast the 9's from the dividend, setting 
the excess to the right, also; if both excesses on the right be equal, the 
work is presumed to be correct. 


Norrs.— 1. Should the excess from the divisor be 0, it is evident we need 
not go over the quotient, as the result found by multiplication would also 
he 0. The excess from the remainder, if any, in that case, will be the same 
as that from the dividend; and if there be no remainder, the excess from 
the dividend must be 0. 

2. The divisor, 996, and its corresponding dividend, might be taken in 
proving the work in the foregoing example, but in that case, the correspond- 
ing remainder, 800, must be taken, instead of 25, 300 being the true re- 
mainder for 996. Try it, the excess will be 0. 

8. It may be well to remark, also, that when the division is continued into 
decimals, the true remainder, and not the decimal, must always be taken when 
using this method of proof. 

4, It is hardly necessary to say that, should there be a misplacing of figs 
ures (an occurreuce, however, which is very rare), this nvethod of proof will 
fail, as it is evident the swms of the digits will be the same, regardless of their 
local positions. 


When the division is performed by the simplified methods, the 
proof can be obtained by the short methods for multiplication 
commencing at page 71; as illustrated in the two following examples: 


Exam. 1. In 497786 square inches, how many square 
reer. 


To divide by 144 (sq. inches in a sq. foot) - 
both numbers are multiplied by 7, giving 


34841521008. 3484|152 + 1008 
The quotient is 8456, and the remainder, 504, ee ere 
To prove the work, 8456 is multiplied by 1008, 3456 eS 


and the remainder, 504, is added. This is per- 504 
formed by simply setting 8 times 3456, or 27648, 27648 
three places to the right, under the remainder, “3484152 
and adding; this gives 8484152, the dividend. 

(See example 2, page 71.) 


On Smee Drvisors. 65 


Exam. 2. Divide 3174473 by 497. 


Doubling both numbers, we have 6348946 to be divided by 994, 
a simple divisor. The quotient is 6387, and the remainder, 268. 


To prove the work, 6387 is multiplied by 994, eyeey + 994 


and 268 added. The process is performed by 38 se 
setting 6 times 6387, or 38322, three places to the 6387968 
right, under the remainder, and subtracting; this 33/999 
gives the dividend. (See example 3, page 72.) $348946 


Nots.— The proof is to be always taken before reducing the remainder toa 
decimal. 


ON SIMPLE DIVISORS. 


Divisors may be simplified by any process that will make them 
10, 100, 1000, 10000, etc., or that will make them a little more, ora 
little less, than these; as, 101, 102, 1003, 10007, etc.; or 91, 92, 98, 
etc.; 991, 992, 993, 9989, 9999, etc. 


Norge. — Composite numbers, when not too large, can be readily divided by 
using their component factors. This method is called successive division. A 
composite number is one that may be produced by multiplying together two or 
more numbers. Thus: 18 isequal to6x3; or 9x2; or3x3x2, 


Exam. In 12872 cubic feet of earth, how many cubie 
yards, or loads ? 


In this, the component factors of 27 (cub. feet in a 


cub, yd.) are 3 and 9 (3 X 9=27). We divide first, : ate ms 
by 3, and the result by 9, to get 476.74 cubic yards, aie Fil 


or loads. 


Nore. —Jn making calculations where the divisor is constant, and is 
a composite number, the desired results are more easily obtained by 
successive division than by the usual long methods; thus: 2240 
pounds toa gross ton, the factors are 40, 7 and 8 (40 & 7 & 8 = 2240). 
Again, 5289 feet to a mile; the factors are 60, 8 and 11 
(60 « 8 & 11 = 5280); and 43560 square feet to an acre; the factors 
are 40, 11 and 99 (40 « 11 & 99 = 43856V) ; and by the short methods 
already established, divison by 99 is the simplest part of the 
process. (See page 5d.) 


5 


66 On Simpie Divisors. 


The following suggestions will aid the student in 
obtaining simple divisors : 


To divide by 
114: 11.5, or 115, multiply by 8; the results are 92 and 920, simple 
divisors. . 
124: 12.5, 1 25, 125 or 1250, multiplied by 8 will give 100, 10, 1000 
and 10000. 
13: 184, 14.25, 1.325 or 1825, multiplied by 8 will give simple divisors. 
13}: 13.5, 1.85, 185 or 1850, multiplied by 8 will give simple divisors, 
14: The component factors are 2 and 7; or multiply by 7 the result 
is 98. 
144: 14.25, 142.5, 1425 or 14250 multiplied by 7 give simple divisors, 
144: 14.5, 145, 1450, etc., multiplied by 7 give simple divisors. 
To diviace by 14} or 14.25; multiply by 7, the result is 993; use 100 
for the approximate divisor, and add for the quarter, etc. 


Exam. Divide 109185 by 145.. 


In this, both numbers are multiplied by 7, and 
we have 764295+1015. Buta slight inspection 


shows that this can be still further simplified. 109185 + 145 
By setting 145 one place to the right, under ae aS. 


1015; and the given dividend one place to the 75313765 10005 

right under the new dividend, and subtracting 3765 

in each case, we get 10005 for a simple divisor. aaa 

The required quotient is 753. 

164: 16.5, 165, 1650, 16.6, 166, 1666, etc.; 167, 16.7, 1670, 1675; 
168, 16.8, 1680, etc.; and 163, all multiplied by 6 give simple 
divisors; thus: 164 x 6 = 99; ete. 

17: Multiplied by 6 gives 102; and 174, or 17.28x4=—69. (See 
pages 250 and 251.) ; 

174: 17.5, 175, 1750, etc.; multiplied by 6 give simple divisors. 

17$: 17.75, 177.5, 1775, 17750, etc.; multiplied by 8 x 7 give simple 
divisors, Thus: 17? or 17.75 x 8 = 142, and 142 x 7= 994, 
a simple divisor. . 

18: The factors are 3 and 6; or 18 X 6 = 108, a simple divisor. 

18}: 18.25, 1825, 18250, etc.; multiplied by 4 give 78, 730, 7300, 
73000, etc.; and the teeta of dividing by these is given at 
pages 52 aod 53. 


On Srmpce Drvisors.: 67 


183: 18.75, 1875, 18750, etc.; multiply by 4and to the result add } 
of itself. Thus: 183 x 4= 75; 4 of which is 25, and this 

added to 75 gives 100; etc. 

19: Multiplied by 5; the result is 95: 194, 19.25 or 1925 multiplied 
by 5 and a fifth of the number itself added; thus: 
1925 k 5 = 9625 + + of 1925 = 10010. . 

It must be carefully borne in mind that, whatever change is made 
in the divisor, to simplify it, a similar change must be made in the 
dividend to preserve the relation; and that when the divisor can be 
simplified, any multiple, sub-multiple, or aliquot part of it can also 
be simplified. (See page 250.) 

And now, if the numbers from 19 to 100, be taken and examined, 
the subject will be found not only interesting and instructive but it 
will be found that division by the majority of these, their multiples 
and sub-multiples, will be exceedingly simple. For instance, most 
of the twenties multiplied by 4; the thirties by 3; the forties and 
fifties by 2, etc., will give simple divisors. 

A few more examples of a practical nature, before closing this 
chapter, may be found helpful. 


Exam. In 1269 pounds of oil; how many gallons, 74 


pounds to the gallon ? 


In this, we simply add athird; and atenth of the j 
result is the number of gallons, and the decimal] of 439 i 
a gallon. In other words, by adding a third to 74 it 172.810 
becomes 10, a simple divisor. 


Exam. In 12134 feet; how many perches of 164 feet ? 


12134+163 
Here, both numbers are multiplied by 6, and we 7281/04 99 
have 72804+99. The answer is 73533, or reducing 7/28 
the fraction to a decimal, .3939, ete. pea U 
73539 . 


Exam. In 34682487 feet; how many miles of 5280 


feet? 

We give the solution of this two ways: 

First: By using the factors (60 x 8 xX 11=5280). 3468248 /7_ 
Cutting off one figure, and dividing by 6, divides Yee pal 
by 60. Then an eighth of this, and one eleventh —$56816528 
of the eighth. 


«668 On Smmpre Divisors. 


Second: Here, we take the component fac- 
tors, 80, 11 and 6, or 80 and 66 (66 « 80 = 5280) 
and divide, first, by 80, by cutting off one 
figure and dividing by 8. ‘hen, to divide by 
66, we add half to get 99, a simple divisor. 
The remainder of the process is clear. 

The answer is: 6568.6528 miles. The deci- 
mal may be carried to any desired length, by 
annexing ciphers to the remainder and divid- 
ing by 99. 


Exam. In 76824763 square feet; 


3468248 .7 


433031 .0875+66 


216765 .5487 33 

~ 6502/96. 6312-99 
65)02 
66 


6568/6463|12 


64/63 
64 
.6528)39 


how many acres, 


true to five places of decimals, 43560 sq. ft. to the acre ? 


The component factors of the divisor, in this, are 40, 11 and 99 
*(40 x 11 x 99 = 48560). The solution is left for the student. 
Should he fail to see his way clear, he is referred to page 59, where 


the explanation is given. 


And when the digits follow in their natural order, the division can 


be simplified, as illustrated in the following: 


Exam. Divide 15241578750190521 by 123456789. 


15241578750190591 + 193456789 
1219392630001524168 987654319 


123456787|87654322|01 + 99999999]09 
1/23456787 
2 


123456789/1111111101 
PAVIA OL 


Here, we first multiply both dividend and divisor by 8, setting 
the product under each respectively, one place to the left of units, 
and adding, in both cases, we take the results for a new dividend 


: 
; 
j 
4 
: 
. 
; 
3 


On SIMPLE DIVISORS. 69 


and new divisor. Cutting off 09 from the divisor, now, and 01 
from the dividend, we divide the remaining part of the dividend 
by the remaining part (99999999) of the divisor, 1 being the key, 
and we get 123456789 for quotient, and 11111111 for remainder, 
to which 01, cut from the dividend, is annexed, and the re- 
mainder is then 1111111101. 

The result thus obtained is too large, being the quotient for 
9999999900; to correct, we subtract 9 times (09 cut from the 
divisor) the quotient, 123456789, or 1111111101. There is no re- 
mainder, 123456789 being the quotient. 


Notes.—1. Multiplying the terms by 8, setting the products one place to 
the left, and adding, we need scarcely observe, is multiplying by 81. 

2. The complement of 9999999909 being 91, the quotient may be obtained 
without cutting off the figures as was done in the example, by adding 91 
times the quotient. In that case there would be no subtraction. And on 
examination, it will be found that is what we have actually done, because 
to multiply by 91, we have used the short method, namely, multiplied by 
100, and subtracted 9 (100 -—9= 91). Hence, the 


Rue. To divide by the nine digits in their natural order : 

Set 8 times the dividend under itself one place to the left, and 
divide the sum by 9999999909, or by eight 9’s, 09. 

And when the digits are in a reversed order, we would proceed as 
in the following: 


Exam. Divide 975461057789971041 by 987654821. 


975461057789971041 + 9876543821 
78036884623 19768328 7901234568 


8)7901234568/0987654321 + 80000000001 


987654321 |0123845679V125 
0123456790125 


In this, we multiply the terms by 8, setting the products one 
place to the left, as in the last example, and adding, the new di- 


visor is 80000000001, the key being 4. 


ROCF. On SIMPLE DIVISORS. 


Dividing, first, by 80000000000 (simply cut off ten figures from the 
right of the dividend, by the vertical line, for the ten ciphers, and 
divide by 8), we get 987654321 for quotient and |01234567890125 


for remainder. From the result thus found we subtract 4 of the 


quotient, set in proper position; that is, 1-80000000000 part of 


987654321, or .01234567890125. There is no remainder, 9876543821 
being the required quotient. Hence, the 
Rue. 70 divide by the nine digits written in reversed order: Set 


8 times the dividend under itself one place to the left, and divide — 


the sum by 80000000001, or by 8 followed by nine ciphers and 1. 


Nors. — To get 1-80000000000 of the quotient, 987654321, is to divide the 
latter by 80000000000, and to do so we simply cut off the ten ciphers and 
divide by &: But we find that the number to be divided (the quotient) does 


not contain ten figures to be cut off, to correspond with the number of: 


* ciphers, so we prefix a cipher to supply the deficiency; the number to be 
divided then is .0987654821, and the eighth part of this is .01234567890125, 
which is equal to the remainder from which it is subtracted, giving 0 for 
remainder (all of which can-be seen at a glance after a little praptive with 
our methods). 


Mixep Noumpers. 


A Mixed Number consists of a whole number with a 
fraction annexed; as, 344, 1064, 2494, etc., and division 
by these will be found as simple as viet numbers 5 
thus : 


To divide by 2494, or tts equal, 249.25: Multiply by 4 and the 
new divisor is 997, a simple divisor (not forgetting to multiply the 
dividend, also, by 4). 

2514, or its equal, 251.50: Multiplied Bye 4, gives 1006 for a sim- 
ple divisor. 

1672, or its equal, 167.8383, etc. (3 being repeated), whick may 
be written 167.84: Multiplied by 6, gives 1007 for a simple divi- 
sor; and so of other mixed numbers. 


ea had 


SHORT METHODS FOR MULTIPLICATION, 


Before entering into the more important of our short 
methods for Multiplication, we deem it proper to give, at 
the commencement, a-few of the more simple, with which, 
no doubt, some of our readers are already familiar, but, a 
knowledge of them is essential to all, to fully understand 
the several cases which follow. 


To multiply by 1, followed by any number of ciphers: 
‘Simply annex to the multiplicand as many ciphers as there are in 
the multiplier. 


Exam. 1. Multiply 475891 by 1000. 
475891000 


Here, there are three ciphers in the multiplier, and we simply 
annex three ciphers to the multiplicand to find the product, 
475891000. 


Exam. 2. Multiply 475891 by 1007. 


475891... 
9331237 


479229237 


2 SHoRT METHODS FOR MULTIPLICATION. 


Here, we first multiply by 1000, as in example 1, using periods, 
or dots, instead of ciphers, and to the product add 7 times the 
multiplicand. 


Exam. 3. Multiply 475891 by 993. 


4(DB91 Ss 
3331237 


472559763 
Here, we first multiply by 1000, as in the two previous exam- 
ples, and from the product subtract 7 times (1000 — 993) the 
multiplicand. 
To multiply by 21, 31, 41, etc.: Simply set the product by the 
tens under the multiplicand, in proper position, and add, thus: 


Exam. 4. Multiply 6853 by 71. 


68538 x 71 
47971 


482563 


And if ciphers come between the two digits of the multiplier, 
proceed in the same way, only move the product as many places 
to the left as there are figures in the multiplier; thus: 


6853 x 7001 
A971... 
47977853 
Composite numbers which can be readily factored should 
always be used, so as not to need addition in the multiplication; 
thus: 


Exam. 5. Multiply 3684 by 42. 
3684 x 42 


22104 
154728 


SHort METHODS FOR MULTIPLICATION. Tou 


Instead of first multiplying by 2 and next by 4, and adding the 
partial products, we prefer to use the component factors 6 and 7 
(6 x 7 = 42). Multiplying first by 6, we get 22104, and multi- 
plying this in turn by 7, we have the product, 154728, without 
addition. 


As we shall make use of subtraction to a large extent 
in the short methods for Multiplication, we may be per- 
mitted here to make a slight digression to say a few 
words on 


SUBTRACTION. 


The usual method in Subtraction is to place the less number 
below the greater, with units under units, etc.; but it will be 
often found of greater advantage to reverse this order, by having 
the less placed above the greater; or, by having both the less and 
the greater contained in one number. Both the latter methods 
will be used in the following short methods for Multiplication, 
but in the one case the words upper and lower will be interchanged 
throughout the rule, and in the other we shall point out the two 
numbers when contained in one. 


Exam. 6. Multiply 6847 by 4193. 


42.. 

6847 x 4193 

47929 7 
Q7T5T4.. 
28709471 


Here, instead of multiplying by 4193 and adding the four par- 
tial products, we take 4200 for approximate multiplier. The dif- 
ference of the multipliers is 7, by which we first multiply. getting 
47929. Now, we see that 7 is one of the factors of 42; dividing 


v4 SHort METHODS FOR MULTIPLICATION. 


42 by 7 gives the other factor, 6. It is evident, now, that by 
multiplying the product of 7, or 47929, by 6, we get the product 
for 42, or 287574, and that by simply moving this number three 
places to the left of units, it represents the product of 4200. 
Then subtracting the wpper number from the lower, in other 
words, taking the product of 7 from that of 4200, gives the pro- 
duct for 4193 (4200 — 7 = 4198). 

And if the multiplier were 41993, 4199938, etc.; 3493, 34993, 
etc.; 3495, 84995, etc., we would proceed in a similar manner. 
using 42000, 420000, etc.; 38500, 35000, etc., for approximates; 
and so on with other numbers of a like nature. 


Exam. 7. Multiply 4876 by 999. 


4376... x 999 
4371624 


Here, we multiply by 1000, by simply conceiving three ciphers, 
represented by dots, annexed; the result is 4376000, or rather, 
.4376..., as seen in the margin. Now, if from this we take once 
the multiplicand, 4376, the difference will be the product of 999 
(1000 —1= 999). It will now be observed that the product of 
1000, and the product of 1, are both contained in the one expres- 
sion, 4376..., and that by simply taking 4376, the product of 1, 
from the whole, 4376.. , the product of 1000, the difference is the 
required result; the subtraction being performed without setting 
down 4876 a second time. 


REMARK.— It may, perhaps, be well to remark, that it is imma- 
terial in Multiplication which factor is taken as multiplier,. or 
which as multiplicand; or what position the partial products hold 
with reference to the multiplicand, whether they be placed to the 
right or the left; provided they are assigned the proper posi- 
tion with respect to each other. 


SuHort MrtTHops FOR MULTIPLICATION. V5 


Exam. 8. Multiply 3984 by 3476. 


cae 7 
3984 x 3476 
16 138904... 
55616 
13848384 
A short inspection of the factors here, shows that by taking the 
multiplicand, 3984, as the multiplier, the process can be shortened. 
Taking 4000 for approximate, the difference of the multipliers is 
16. Multiplying 3476 by 4, and annexing three ciphers (dots) 
gives the product of 4000. Now, we observe that 4, the signifi- 
cant figure of 4000, is contained 4 times in 16, the difference of 
the multipliers, and that by multiplying 13904, the product of 4, by 
4 (found by dividing 16 by 4), and setting the result, 55616, in 
proper position, it represents the product of 16. Subtracting 
this, now, from 13904...; that is, taking the product of 16 from 
that of 4000, gives the product for 3984 (4000 — 16 — 3984). 


To multiply by any number from 11 to 19. 
Exam. 9. Multiply 743586 by 11. 


743586 x 11 
8179446 


Here, we simply set down first the unit figure, 6, of the multi- 
plicand, then add the figures from right to left, carrying when 
necessary as we proceed, thus: 6 and 8 are 14; 8 and 5, and 1 
carried, are 14; 5 and 3, and 1 carried, are 9; 3 and 4 are7; 4 
and 7 are 11; 7 and 1 carried are 8. 


Exam. 10. Multiply 7468 by 17. 


7468 x 17 
126956 


76 SHorRtT METHODS FOR MULTIPLICATION. 


Here, we multiply by 7, the unit figure of 17, adding the figures 
of the multiplicand as we proceed, thus: 7 times 8 are 56; 7 times 
6 are 42, and 8 (the unit figure) and 5 (carried) are 55; 7 times 4 
are 28, and 6 (the figure to the right of 4) and 5 (carried) are 39; 
7 times 7: 49 and 4, and 8 (carried) are 56; 7 (the last figure) and 
5 (carried) are 12. 


Norte. — In such cases it is better to always add the figure of the multipli- 
cand first, adding in the figure carried after. 


Exam. 11. Multiply 7258 by 1013. 


[208.21 Mae 
94354 
7352354 
Here, we first multiply by 1000, and next by 18 (short pags 
and add. 


Exam. 12. Multiply 13986 by 3684 


AE: Fass 

3684 x 13986 

SIS (6S 1k 
51524424. 


Taking the multiplicand in this (seeing that it is near 14000) 
for the multiplier, the process can be shortened at once. Multi- 
plying 3684 by 14 (short method), and then annexing three 
ciphers (dots) we have the product of 14000, or 51576... Now, 
the difference of the multipliers is also 14, and it is evident that 
if 14 times the multiplicand, 3684, be taken from the product of 
14000, the difference will be the product of 18986. Now, the 
product of 14, and also of 14000, are contained in the one expres- 
sion, 51576...; all we have to do, then, is to subtract 51576 from 
51576..., and the difference is 51524424, the required product. 
(See example 7.) 


Suort Metruops FoR MULTIPLICATION, 7? 


Exam. 13. Multiply 35982 by 3286. 


36... 
35982 x 3286 
18 59148 
118296... 
118236852 


Here, we see tha. 35982 is near 36000, so we take these two as 
the multipliers. Their difference is 18. Multiplying 3286 by 18 
(short method), the product is59148. Multiplying this by 2 gives 
the product of 36 (18 x 2 = 36); annexing three ciphers (dots) 
gives the product of 36000, or 118296... Now, if from 36000 we 
take 18, the difference is 35982. Subtracting 59148, the product of 
18, from 118296..., the product of 36000, gives 1182386852, the 
required product. 


Exam. 14. Multiply 46782 by 27985. 


Oa 

46782 x 27985 

654948 45 
1309896... 
1309194270 


If the difference of the multipliers, in this example, was 14 in- 
stead of 15, wecould proceed as in the last. Here, we set 14 
times (short method) the multiplicand under itself, and we see at 
a glance that if both numbers be added the result is the product 
of 15 (14 + 1). We donot add, however, but set 2 times 654948, 
the product of 14, four places to the left, having first drawn a 
separating: line, and conceiving three ciphers annexed, we have 
the product of 28000, or 1309896... Now, if from this product, 
the sum of the two numbers immediately above it be taken, the 
difference will be the product of 27985 (28000 — 14 + 1). 


78 SHort METHODS FoR MULTIPLICATION. 


Notre.—It is hardly necessary to say to the arithmetical student, that, in 
taking the two partial products which go to make up the product of 15, from. 
that of 28000, the addition and subtraction go hand in hand, thus: Com- 
mencing at the top; 2 and 8 are 10, from 10, and 0. Carry 1 to 8, 9 and 4 
are 13, from 20 (in this case) and 7; 2 and 7, 9, and 9 are 18, from 20 and 25 
2 and 6, 8, and 4 are 12, from 16 and 4, etc. 


Exam. 15. Multiply 24975 by 5487. 


24975 x 5487... 
135925. 
185789075 


Seeing that 24975 is near 25000, we take these two numbers for 
multipliers; we see by inspection that the difference is 25. We 
multiply by 25000, thus: Conceiving two ciphers annexed to 
5437 multiplies that number by 100, and taking one-fourth of the 
result gives the product of 25, or 135925. To this we annex three 
ciphers (dots), and we get 185925..., the product of 25000. Now, 
taking 25 from 25000, gives 24975, that is, 135925... minus 135925 
(both contained in the one number), gives 135789075, the required 
product. 

When the quotient obtained by dividing the difference of the multt- 
pliers by the significant part of the approximate multiplier, consists 
of two or more figures which can be readily factored, we would pro- 
ceed as in the following: 


Exam. 16. Multiply 59832 by 78481. 


Oe ae 
59832 x 8431 


168 98 470586. ...1882344 
13176408 


4692683592 
Taking 59832 and 60000 for the multipliers, in this example, 
we find cheir difference to be 168, which contains 6, the signifi- 


SHort Metuops For MULTIPLICATION. 19 


cant part of 60000, 28 times. Multiplying 78431 first by 6, and 
annexing four ciphers (dots) we have the product of 60000. Now, 
28 times 6 is 168, and if 168 be taken from 60000 it leaves 59832; 
in other words, if 168 times 78431 be taken from 60000 times that 
number, or 470586...., which we already have, the difference 
will be the product for 59832. To get 168 times the multiplicand, 
78431, we take 28 times the product of 6. To multiply by 28, we 
use its component factors, 4 and 7 (4 X 7 = 28), setting 4 times 
470586 a little to the right, as shown in the margin; then 7 times 
that product, or 13176408, is set under the product of 60000 and 
subtracted, giving 4692683592, the required product. 


When one part of the multiplier, or of the multipli- 
cand, 7s a multiple of another part, illustrated in the 
following: 


Exam. 17. Multiply 47634 by 16128. 


47654 x 16128 
762144... 
6097152 
768241152 


A glance at the multiplier, 16128, in this example, shows that 
128, the three last figures, is 8 times 16, the two first, in other 
words, 128 is a multiple of 16. 

Multiplying first by 16 (short method), and conceiving three 
ciphers annexed to the result, gives the product of 16000; then, 
‘setting 8 times 762144 in proper position, and adding, we get the 
required product. 


Nore.—If the muitiplier or the multiplicand had been 12816, we would 
proceed in the same way, first multiplying by 16, as in the example; then 
setting 8 times the product of 16, three places to the left and adding. And 
if one of the factors were 1612832, 3212816, 1283216 or 1632128, the process 
would be equally simple; multiplying first by 16, then 8 times the product 


80 SHortT METHODS FOR MULTIPLICATION. 


of 16, and next, 2 times the product of 16; taking care to place the partial 
products in proper position with respect to one another. And so of other 
numbers similarly combined; such, for instance, as 812, 8012, 815, 3015, 
etc.; 412, 416, 424, 4024, etc.; 642, 6042, 756, 7056, etc., etc. 


Division REVERSED. 


Many extraordinary contractions in Multiplication may 
be obtained by simply reversing our new methods for Di- 
vision. A few examples will illustrate. 


Exam. 1. Multiply 756 by 334. 


TOOT: 
1512 


252504 


Here, we simply annex three ciphers (dots) to 756, and under 
the result set 2 times 756, or 1512. The sum of these two num- 
bers, without actually setting down the result, is evidently 757512, 
and we divide both numbers, considered as one, by 3, getting 
252504, the required product. 

Reason: To divide by 334, we would multiply by 3, getting 
1002; divide first by 1000 and subtract 2 times the quotient. In 
the multiplication we reverse the process, multiplying first by 
1000, adding 2 times the multiplicand, and then dividing by 3. 

And if one or both of the factors were 3334, 83384, etc., the 
process would be equally simple. 


Exam. 2. Multiply 1668 by 478. 


4783894 
797304 


Here, we simply set 8 times 478, or 3824, to the right and di- 
vide by 6, and we have 797304, the product. 


SHort MetuHons FoR MULTIPLICATION. 81 


Reason: To divide by 1668, we would multiply it by 6 (being 
nearly the sixth of 10000), getting 10008 for new divisor. 

To divide by 10008, we first divide by 10000, then subtract 8 
times the quotient. To multiply by 1668 (which in the example 
we have made the multiplier), we reverse the division by first 
multiplying by 10000, adding 8 times the multiplicand, and then 
dividing by 6. 

And if one or both factors were 16668, 166668, etc., the process 
would be equally simple. 


Exam. 8. Multiply 7342 by 8334. 


61188228 


Setting 8 times the multiplicand, or 58736, five places to the 
right, in this, and looking on both numbers as one (that is, as 
being added), we divide by 12 to get 61188228, the required 
product. 

Reason: To divide by 8334, we multiply it by 12, getting 
100008 for divisor; reversing the division, we have the multipli- 
cation. 


Exam. 4. Multiply 142858 by 37684. 


142858 ) 37684...... 
1000006 226104 
5383460872 


In this we take 142858, as the multiplier: To divide by this 
number we multiply it by 7, getting 1000006 for divisor, as shown’ 
in the margin. 

To multiply by 142858, we reverse the division by multiplying 
37684 by 1000000, adding 6 times 37684, or 226104, and dividing 
by 7. 


\ 


aa 
fas 


82 SHort METHODS FOR MULTIPLICATION. 


And the numbers 1428, 1429, 14286, 14287, 142857, etc., will — 
be found equally simple, and so with other numbers, 


Exam. 5. Multiply 123456789 by 123456789. 


198456789.......... 
11234567799 
9)1934567878765432201 
9)13717420875 17114689 
15241578750190521 


In this we simply conceive ten ciphers annexed to the multipli- 
cand, and from the result we take 91 times (short method) the 
multiplicand, or 11234567799; then dividing the difference by 9 
and 9, in succession, we have the required product. 

This is the reverse of division by the nine digits in direct order, 
(See example, page 68.) 

Ruue. Zo multiply any number by the nine digits in direct order: 
Annex, or conceive to be annexed, ten ciphers to the number to 
be multiplied; from the result take 91 times the said number and 
divide the difference by 9 and 9 in succession. 


Exam. 6. Multiply 987654321 by 987654321. 


U8iGo482 1 chaste 
79012345680987654321 
8779149520 1097389369 
9754610577899 7 LO4L 


Conceiving ten ciphers annexed to the multiplicand, in this; 
multiplying the result by 8, and adding the multiplicand, we 
divide the sum by 9 and 9, in succession, as in the preceding 
example, and we have the required product. | 

This is the reverse of division by the nine-digits written in re- 
versed order (see example, page 69). 


SHort METHODS FOR MULTIPLICATION. 83 


Rue. To multiply any number by the nine digits in a reversed 
order: Annex, or conceive to be annexed, ten ciphers to the 
multiplicand; to 8 times the result add the multiplicand, and 
divide the sum by 9 and 9 in succession. 

Nore.— An unlimited number of such examples could be added, but, for 
the student who has carefully read the preceding pages of this work, enough 
has been given by way of suggestion, 

To square any number of two digits we give the following simple 

Rowe. (1) Add to the given number the difference between it 
and the next higher number ending with a cipher. (2) Subtract 
from the given number the number which was added. (3) Multi- 
ply the sum and difference thus found, and to the product add 
the square of the figure added. 


Exam. 1. What is the square of 47? 
First, add 38, making ..... 50 
Second, subtract 3, making 44 


Their product is......... 2200 
Now add 3 times 3, or... 9 


2209, the square. 
Or, if more convenient, subtract from the given number the 
difference between it and the next Jower number ending with a 


cipher; then add to the given number that which was subtracted, 
and proceed as above, thus: 


Exam. 2. What is the square of 91? 


First, subtract 1, making. . 90 
Second, add 1, making... | 92 


(ete eee 


Their product is......... 8280 
AMS Xl OFime wad A 1 


eee 


8281, the square. 


84 SHortT METHODS FOR MULTIPLICATION. 


The process is founded on the principle that: Zhe product of 
the sum and difference of two numbers, is equal to the difference of 
their squares : 

If 47 + 3 be multiplied by 47 — 3, the product 1s equal to 4% 
squared minus 8 squared; that is, 


(47 +8) x (47 —8) = 47733 


Now, 47 + 3 is the same as 50, and 47 — 3, the same as 44; 
therefore 50 x 44 is equal to (47 x 47) minus (3 x 3), that is, 
when 50 is multiplied by 44, and the square of 3, or 9 added to the 
product, the result will be the square of 47; thus: 


50 x 44 — 2200; and 47? 3? = 2209 — 9; that is, 
9900 — 2209 — 9 


Now, it is a well-known axiom that: if equals be added to equals, 
the sums will be equal. If, now, 9 be added to the two last 
expressions, which are equals, the sums will be equal, thus: 
2200 + 9 = 2209 —9+9; subtracting 9 and adding it at the — 
same time leaves the last expression 2209; and adding 9 to 2200 
makes it 2209, which is the square of 47. 


Norz.— A little practice will enable a person to readily square any number 
consisting of two digits mentally by this simple method. 


This method will be found convenient in many cases 
where the number to be squared consists of three figures. 


Exam. 3. What is the square of 1092 


109? = 100 x 118 + 81 = 11881, at sight. 


Numbers consisting of three figures may be readily squared on 
the same principle, thus: 


SHorT Metuops For MULTIPLICATION. 85 
Exam. 4. What is the square of 432? 


400 x 464 = 185600 


459? == 4 30 x '34'== * 1020 
Q2x 29= 4 
186624 


Subtracting 32 gives 400, adding 32 gives 464; the results are 
multiplied, giving 185600. The square of 32 is added next; sub- 
tracting 2 gives 30, adding 2 gives 34; the results are multiplied, 
giving 1020. Finally, the square of 2 is added, making 186624, 
the square of 4382. 


Exam. 5. What is the square of 371 ? 
442 x 300 — 152600 


Bila + 12%. 10:—+. 5040 
1x L= 1 
137641 


Adding 71 makes 442, subtracting 71 makes 300; their product 
is 132600. 

Adding 1 to 71 makes 72, subtracting 1 makes 70; their pro- 
duct is 5040. 

Adding the square of 1 gives 137641, the square of 371. 


Rute. To square any number of two figures: (1) Multiply the 
units by the units and set down the unit figure of the product. (2) 
Multiply the sum of the units by a single figure of the tens, set down 
the unit figure of the product. (8) Multiply the tens together to com- 
plete the square, carrying as usual. 


Exam. What is the square of 74? 
Ans. 74=74 X 74 = 5476. 


Here, say 4 times 4 are 16; set down 6, and carry 1; then, 7 times 
8 (4 + 4) are 56 and 1 are 57; set down 7, and carry 5: next, 7 times 
7 are 49, and 5 are 54 completes the square. (See page 271.) 


86 Saort Mergops ror MOULtIPLicatTIoN. 


Rue. To square numbers of two figures ending in 5: (1) Multiply 
the 5's together and set down the result in full. (2) Add 1 to either 
Jigure of the tens and multiply the other by the number thus increased. 


Exam. What is the square of 75? 
Ans. 75°= 15 + 75 = 5625. 
Say 5 times 5 are 25; set down in full; add 1 to either 7 and say 
8 times 7 are 56 to complete the square. 
Reason: 7 xX 75 = 80 xX 70 plus 5 X 5 = 5600 + 25 = 5625. 


The rule is applicable in many cases to three, four or more figures, 
thus: 125. 185, 145, etc., 295; 895; 39995, etc. 


Exam. What is the square of 695? 


Ans. 695? = 695 X 695 = 483025. 


In this say 5 times 5 are 25; set down in full; add 1 to 69 and say 
70 times 69 are 4830 to complete the square. And so with the others. 
and similar nuinbers. 

And the same rule-can be applied to any two figures, whose units 
when added make 10; the tens in both numbers being alike; as 47x 
43; 76 x 74; 58 X 52, etc., and in many cases to three, four, or 
more figures, as 172 x 178; 127 x 123; 196 x 194; 193 x 197; etc., 
292 X 298, 293 x 297; 394 x 396; 4998 x 4997; ete. 


Exam. Multiply 76 by 74. Ans. 76 X 74 = 5624, 
Say 4 times 6 are 24; set down in full; add 1 to either 
7, and say 8 times 7 are 56; for the reason that 76 X 74= 
S80 X 70 plus 6 X 4 = 5600 + 24 = 5624. 

Exam. Multiply 397 by 393. Ans. 397x393 = 156021. 
Say 3 times 7 are 21; set down in full; add 1 to 39 
(either factor) and say 40 times 39 are 1560 to complete 
the product. (See examples on page 87.) 

NotE.— When the units are 1 and 9 the second figure of the product will 
always be a cipher; as: 71 x 79 = 5609, 691 x 699 = 483009. Say 9 times1 are 9, 


then 0, 8 times 7 are 56. Say 9 times 1 are 9; then 0; and 70 times 69 are 4830 
completes the work. 


Exam. Multiply i27 by 124. Ans. 127 x 123 = 15621 
Plus 127 multiplied by 1 = 127 


15748. 


Note, — Multiply as if the units made 10, and add or subtract for the difference. 


emer 
eo 
~~ es 
+) Sica 
Sa bint, eel! 


Sport Mrruops ror MULTIPLICATION, 87 


Hence, in all cases where the two units figures make 10, and the 
other figures are alike, the foregoing rule is applicable, as illustrated 
in the following 


ExAMPLES: 
127 < 123 = 15621: Here, because 3 and 7, the unit figures, make 
112 x 118 10, and 12 and 12 are alike, we simply say 3 
47 X 45 times 7 are 21, setting it down in full; then, 
84 x 86 adding 1 to either 12, calling it 18, we say 12 
134 x 136 | times 13 are 156, which completes the product, 
etc. 15621. 


And tf one or both factors contain a fraction, the process will be 
found equally simple; thus: 


24 x 264 = 636: In this, 6 and 4 make 10, and the 2’s are alike; 


284 x 22 say one half of 24 is 12, tocarry; then, 6 times 4 

36 X 34} are 24, and 12 are 36; set down in full; now add 1 

48 x 424 to 2 and say 3 times 2 are 6; this completes the 
etc. product. 


241 & 261 = 6431: Here, we say } of 24 is 6, to carry: then, 6 


123 x 184 times 4 are 24, and 6 are 30; now, 3 times 2 are 
O2t X 3885 6; making 630, or 24 x 261: to this is added } 
as of 26}, or 134, making 643}. 


Note.— When the unit figures are 1 and 9 the second figure of the product 
will be a cipher; thus: 61 x 69 = 4209, 


By this simple rule, then, such numbers as 21 < 29, 22 x 28, 
23 X 27; 31 x 39, 82 x 38, 85 xX 35; 47 x 43, 48 x 42, 58 x 52; and 
up in the sixties, seventies, eighties, etc., may be multiplied 
together without effort. Also, 191 «199, 192 x 198, 1992 x 1998, 
1202 x 1208, 1807 « 1303, 798 x 792, 6993 x 6997, etc. (See examples, 
page 260 ) 


Exam. What will 198 hats cost at $1.92 each? 


In this, 2 and 8 make 10, and the 19’s are alike. 
Say 2 times 8 are 16; set down in full, then, add 1 to 198 x 1.92 
19 and say 20 times 19 are 380; this makes se 16, $380.16 
the cost. 


88 Meruops oF PRooF. 


First: by casting out the 9’s: cast the 9’s out of both factors, 
as in the proof for Division, page 63, and reserve the excesses. 
Multiply these excesses together and from the result reject 9, also. 
Now, find the excess of 9’s in the product, and if this be the same 
as found from the factors, the work is generally correct. 


In the annexed example, the excesses of the 


factors are 4 and 3; their productis12. Taking 43976... .4 
9 from 12 leaves an excess of 8. Then, reject- 14682. ...3 t 12 
ing the 9’s from the product, the excess is 3, 6353878232 3 


also. (See page 63, and note 4, page 64.) 


Second: casting out 11’s: commence at the units of the multipli- 
cand and add all the digits in the odd places; then, all those in 
the even places, and from the former, increased if necessary, by 11, 
subtract the latter, and reserve the excess. Proceed in like manner 
with the digits of the multiplier, and reserve the excess. Multiply 
these two excesses and take the even from the odd (increased by 11, 
if necessary) and reserve the excess. Finally, in a similar manner, 
find the excess in the product, and if this be the same as that from 
the factors, the work is generally correct. 


In the annexed example, the sum of the digits 43276. ..2 
in the odd places of the multiplicand, is 12, and 14682. ..8 t 16 
that in the even, 10; the excess is 2. Now, in 6353782382. .... 5 
the multiplier, the digits in the odd places make 

9, and in the even, 12. Increasing 9 by 11 we get 20 from which 
12 is taken, leaving an excess of 8. Then, 8 times 2 are 16, and 
taking the even from the odd in this, we say 1 from 6 leaves 5, the 
excess. Or, 11 from 16 leaves 5. Finally, the digits in the odd 
places of the product make 22, and those in the even, 17; the excess 
is 5, also, the same as found from the factors. 


Norg. — In all cases where the odds are less than the evens the odds must be 
increased by 11. Those figures can be applied to prove Addition, Subtraction 
and Division. The 11 is more reliable than 9, and may be used with advantage 
as a check figure to test the correctness of the footings of the Cash Book, 
Journal, etc. By applying both 9 and 11 to the same operation, an almost abso- 
lute certainty of its correctness would be obtained. 


CANCELLATION. 


Arithmetical calculations are frequently abbreviated in 
a wonderful manner by what is called Cancellation, that 
is, rejecting equal factors from numbers bearing to each 
other the relation of dividend and divisor; or, from mul- 
tiplier and divisor in operations requiring multiplication 
and division. 


Exam. 1. Multiply 476 by 312 and divide the product 
by 416. 


476 x 312 


1428.. 
5712 


+ 416 
104 


357 is the required result, found by the methods already estab- 
lished. 


50 CANCELLATION. 


Bringing cancellation to our aid, the same result is 
much more readily found as follows: 


4 8 

416 476  ~—s-_ «28:19 

104 1428 * 104 
B57 


In this, we see at a glance that the multiplier, 312, is divisible 
by 3, giving 104; and the divisor, 416, is divisible by 4, giving 
104 also. Now, multiplying 476 by 312 is the same as multiplying 
it by the factors, 3 and 104. Next, dividing by 416 is the 
same as dividing by the factors, 4 and 104. Multiplying 476 by 
104, and dividing it at the same time by 104, will not change the 
number. Rejecting the common factors, 104, then, we simply 
multiply 476 by the remaining factor, 3, getting 1428, which is 
divided in turn by the remaining factor, 4, giving 357, as found 
by the long process. 


Exam. 2. Multiply 468 by 800 and divide the product 
by 792, 
192 468). . 8he 
4168 
4 
472|72 


Dividing 792 and 800, in this, each by 8, we get 99 for divisor 
and 100 for multiplier. Annexing two ciphers (dots) to 468 mul- 
tiplies by 100; then to divide by 99, according to the simplified 
method, we divide by 100 and add 1-100 as often as possible. 
Now, to divide by 100 we cut off the two ciphers (dots) which 
were annexed to multiply by 100. From this it will be readily 
seen that the position of the vertical line, in all such cases, is to 
the right of units, or through the decimal point. 


2 0 


CANCELLATION. 91 


Exam. 3. Multiply 423 by 14 and divide the product 


by 728. 
728 493 14 
104 846 9 
32 
14 


Dividing 14 and 728 each by 7, in this, gives 2 for multiplier 
and 104 for divisor. Multiplying 423 then by 2 and dividing by 
104 we get 8|14—8.134+. (See p. 17.) 


Exam. 4. Multiply 51 by 73 and divide by 49. 


aU ete O leat (ol: 
98 102 1/46 © 


Doubling 49 and 51, in this, the multiplier is 102, and the divi- 
sor, 98. From what has been said in example 2, the position of 
the vertical sine, here, is to the right of 738. Adding 2 times 73, 
now, placed in proper position, multiplies by 102. Then divid- 
ing by 98, we get 75|96=75.979+. (See pp. 19 and 21.) 

On examining the process it will be seen that we have added 
2-100 of 73, or 1/46; and in dividing by 98 we have added 2-100 
of 74; then 2-100 of 1, or 1/48 and |02. Now, it is evident that 
the process can be abridged, thus: 


49 51 731. 
98 te a 


ae 


992 CANCELLATION. 


Adding 4-100 of 73, adds for 2 in 102, and also for 2 in the | 
division by 98. Then 2 times 2, or 4, is added, which finishes 
the division by 98, and we have 75|96 as before. 


Exam. 5. Multiply 204 by 68 and divide the product 
by 214. 
214 68 204: 
107 1/36 103 
69|36 
4/83 
64/53 
35 
88 


Dividing 214 and 204, each, by 2, gives 107 for divisor, and 102 
for multiplier. The result is 64/88=64.82+. 

In this it will be seen that 2-100, or 1]36, has been added; then 
7-100, or 4|838, subtracted, which is evidently the same as sub- 
tracting 5-100 (the difference), so we simply subtract 
5-100 of 68, or 3/40, as shown in the margin. This 68 


adds for the 2 in 102, and subtracts for 7 in 107 _38)40 
Then 7 times 4 (3 plus the 1 carried in subtracting 64/60. 
40), or 28, is added, to complete the division by 107. 28 
The result is 64/88, as before. 88 


Exam. 6, Divide 748 times 416 by 412. 


412 748 416 
103 T4808 
__ it 
THOT 


Dividing 412 and 416 each by 4, the divisor is 103 and the 
multiplier 104. Now, we have to add for 4 in 104, and subtract 
for 3 in 103; their difference is 1, so we simply add 1; that is, 


" CANCELLATION. 93 


1-100 of 748, or 7/48. This completes the multiplication by 104, 
and subtracts for 3 in the divisor. But as the addition exceeds the 
subtraction, it is evident that 7, which is in reality a part of 748, 
immediately above it, has got to be multiplied by 3 (in 103), and 
the result, 21, subtracted. Taking 21 from 48 gives the remain- 
der, 27, and adding 7 to 748 gives the quotient, 755. 


Exam. 7. What would be the result of taking $32.64, 
416 times, and dividing the result into portions of $31.04 
each ¢ 


3104 416 3264 
“St a is 
eal 
437|43 


Taking $31.04 and $32.64 as so many cents, throws off the 
decimals, and the multiplier becomes 3264, and the divisor, 3104. 
A moment’s inspection, now, shows that each is divisible by 4, 
giving 776 and 816. These, in turn, are divided by 8, giving 97 
and 102. 

Then, adding 5-100 of 416, or 20|80, completes the multiplica- 
tion by 102, and adds for 3, the complement of 97. Next, 8 
times 20, and then 3 times 1 (carried in adding the remainders 60 
and 80), set in proper position, completes the division for 97. 
Adding the several results, we get 437/43. 

Nortr.— The value of the remainder, .43, is found by multiplying it by 4, 
and the product by 8 (the factors by which we divided in the example). The 
product is 1376, evidently cents in this case, or $13.76, which is not large 
enough to make a portion ($31.04). The answer, then, is, $31.04 could be 
taken 437 times, and $13.76 would remain. 


Exam. 8. A hatter exchanged 172 men’s hats which 
cost him $3.35 each, for a number of boys’ hats, at $1.68 
each; how many of the latter did he receive ? 


94 CANCELLATION. 


REMARK.— In simplifying arithmetical operations, the relation 7 


which the numbers sustain to each other, must be carefully borne 


in mind. If, for instance,the numbers sustain to each other the 


relation of multiplicand and multiplier, whatever operation is 
performed on the one, to simplify the process, the reverse is 
performed on the other, that the product may not be changed; 
and if the relation be that of dividend and divisor, whatever 
operation is performed on the one, the same is performed on the 
other, that the quotient may not be changed. | 

To solve the above example, we multiply $3.35 by 172, and 
divide the product by 1.68; thus: 


6 8 6 
168 335 172 
1008 1005 1032 

“344... 

1720 

845/720 

2/760 

342/960 

24 

984 


Here, 885 and 172 bear to each other the relation of multipli- 
cand and multiplier, and 168, that of divisor to each of the others, 
or, to their product. 


To divide by 168 we multiply it by 6, getting 1008, a simple 


divisor; and multiplying 172 also by 6 we get 1032. Next, to 
multiply 335 by 1032, we take the former for multiplier and mul- 
tiply it by 3 to get 1005, a simple multiplier. Multiplying 335 by 
3, divides 1032 by 3, giving 344. 

The terms now are 1008, 1005 and 344. Multiplying 344 by 
1005, and dividing the product, 345720, by 1008, according to 
our established methods, we get 342/984=342.976. 


CANCELLATION. 95. 


Bringing Cancellation to our aid, we can still further simplify 
the foregoing process, thus: 


168 335 172 


1008 1005 ~ B44)... 
1/032 

342/968 

16 

984 


It will be observed that, in the first process, 172 was multiplied 
by 6, and the product divided by 3, which, by cancellation, is 
simply multiplying by 2. Here, then, we simply multiply 172 by 
2, getting 344. Next, adding 5 and subtracting 8 is subtracting 
3, in this case 38-1000. So we subtract 3 times 344, placed in 
proper position, or 1|032. Now, since the subtraction exceeds the 
addition, it is evident that 1, or rather 2 (1 plus the 1 carried in 
subtracting), has yet to be multiplied by 8 (in 1008), and the pro- 
duct, 16, added to complete the division by 1008. The result is 
342/984, as before. 


Exam. 9. How many pairs of men’s shoes at $3.34 a 
pair can be got in exchange for 346 pairs of women’s, 
worth $2.494 a pair? 


334 9494 346 
1002 998 1088 
95915. 
1/037 
958/463 


To solve this, $2.494 is multiplied by 346, and the number ow 
times $3.34 is contained in the product is the required number. 
Multiplying 334 cents by 3 gives 1002 for a simple divisor, ama 


96 CANCELLATION. 


nultiplying 2494 cents py 4 gives 998 for a simple multiplier. 
Multiplying 334 by 3 multiplies 346 also by 3, giving 10388; and 
multiplying 2494 by 4 divides 346 by 4; or, what amounts to the 
same thing, 1038 is divided by 4, giving 259.5. The terms, now, 
are 1002, 998 and 259.5. To multiply by 998 we use 1000 and 
subtract 2 (that is, 2 times the multiplicand), and to divide by 
1002 we use 1000 also, and subtract for 2; in other words, we ~ 
subtract 4-1000 of 259, or 1/0386. 

In multiplying by 998, however, it must be borne in mind that — 
259.5 has been multiplied by 2, while in dividing by 1002, only 
259 has been multiplied by 2. In multiplying 259 by 4, therefore, 
1 is added (found by multiplying the decimal .5 by 2, the comple- 
ment of 998), making 1|037 to be subtracted; the result is 
258)46 pairs. 


Exam. 10. How many acres of land at $32% an acre 
can be got in exchange for 176 acres, worth $48.50 an 
acre ? 


3 p) 
B94 176 48.50 
7 88 97 

264 


Multiplying 324 by 3, gives 97 for divisor— multiplying 48.50 
by 2, gives 97 for multiplier, also; and, being common factors, 
we reject both. Now, multiplying the divisor, 324 by 3, multi- 
plies the dividend, 176, also, by 3; and multiplying 48.50, one of 
the factors of the multiplication, by 2, divides the other factor, 176, 
by 2. So we simply divide 176 by 2, and multiply the result, 88, 
by 3, to get 264 acres, the required number. Or, add 88 to 176, 
because multiplying by 3 and dividing by 2 is multiplying by 1}. 


Nors. — Other examples, snowing methods of abbreviation, might be — 
added without limit, but those which are given will be sufficient by way of 
suggestion. 


RULE OF THREE. 


“The Lule of Three” (so called because there are al- 
ways three numbers given to find a fourth) has reference 
to that part of Simple Proportion usually taught in 
arithmetic, to find a fourth proportional to three given 
numbers. 


The resolution of this problem is the most important result of 
the theory of Proportion. On account of its great utility and its 
extensive application by merchants, accountants and others, it 
has been called the Golden Rule. 

A thorough knowledge of this problem will enable the student 
to understand, with little effort, questions in Percentage, Interest, 
Discount, Commission, Profit and Loss, and other branches of 
arithmetic having the principles of Proportion for their basis. 


Proportion is the equality of ratios. 

Ratio is the relation which one number bears to another 
of the same kind, with regard to size or comparative 
value. 


Thus, since 8 is double of 4, and 10 of 5, the ratio of 8 to 4 is 
equal to that of 10 to5. The four numbers, 8, 4, 10 and 5 are 
proportionals and constitute a proportion, or analogy. 

In this relation they are usually written thus: 

mes + 4:2 10: 5;-or, simply, 8's 4:33 10 3:5. 

< ° 


98 RULE OF THREE. 


The first expression is read, as 8 is to 4 so is 10 to 5; and the 
second, 8 is to 4 as 10 7s to 5, meaning that whatever relation 8 
has to 4, the same relation exactly 10 bears to 5; in other words, 
whatever number of times 8 is greater or less than 4, 10 is as many 
times greater or less than 5. | 

Now, 8: 4is a ratio, meaning that 8 has a certain relation to 4; 
10:5 is also a ratio, and both being equal, they might also be 
written thus: 


8 A085 


and may be read as in the other cases, or, the ratio of 8 to 4 equals 
the ratio of 10 to 5. 

The two numbers, 8 and 4, are the Zerms of the first ratio, the 
first number is called the Antecedent, and the second, the Conse- 
guent ; 10 and 5, are the terms of the second ratio, 10 being the 
antecedent, and 5 the consequent. 

The value of a ratio is the quotient obtained by dividing the 
antecedent by the consequent; thus: the value of 8 : 4 is 8, or 2; 
and of 10: 5, 4,2, or 2. 

Since there are two terms in a ratio, and two ratios in a pro- 
portion, there must be at least four terms in every proportion. : 

The first and fourth terms of the proportion, or the outside num- 
bers, 8 and 5, are called the Hztremes ; and the second and third 
terms, or the middle numbers, 4 and 10, the Means. 

And when four numbers constitute a proportion the product of 
the means will always be found to be equal to the product of the 
extremes. Thus, in the proportion: 


8: 4 32510 5 


4 multiplied by 10 equals 8 multiplied by 5. 

Hence, any three terms of a proportion being given, the fourth can 
be readily found. Thus, if an extreme be wanting: Multiply the 
two means together and divide the product by the given extreme ; the 
quotient will be the required extreme, and if a mean be wanting: 


RULE OF THREE. U9 


Multiply the extremes together and divide the product by the given 
mean ; the quotient is the required mean. 


Exam. 1. What is the fourth term of the proportion 
UNS er 


Solution: 24 x 4 = 96; then, 96 + 16 = 6, the required term. 
The proportion is now 16: 4 :: 24: 6. 
Proof: 24 x 4 — 96, the product of the means, 

16 x 6 = 96, the product of the extremes, 


Exam. 2. Complete the following proportion: 
A ee De ea 


Solution: 16 x 6 = 96; then, 96 + 24 — 4, the second term. 

In this, two extremes are given, and one mean, to find the other. 
The extremes are 16 and 6; they are multiplied together and the 
product, 96, divided by 24, the given mean; the quotient, 4, is 
the other mean. 


From the foregoing principles and illustrations we de- 
rive the following: 


Rue. (General Rule.) To find a fourth proportional to three 
given numbers : 

I. Arrange the three given numbers in a line, in succession, setting 
the one which is of the same kind as the required term, the third in 
order. 

Il. Lf, by the nature of the question, the required term is-to be 
greater than the third term, put the greater of the other terms in the 
second place; but if the required term is to be less than the third, then 
put the less of the other two in the second place. 

Ill. Multiply the second and third terms together and divide the 
product by the first; the quotient wili be the required term. 


100 RuLE oF THREE. 


Notrs.— 1. If the first and second terms be not of the same denomination, 
they must be reduced to such. 

2. If the third term be a compound number it must be reduced to its 
lowest unit. 


Exam. 38. If a person is to receive $840 for 9 months’ 


services, how much ought he to receive for 90 days at | 


the same rate ? 
9 x 80 = AO Oe 840 
Bit ieee Bee 


The answer must be money, so we put $840 for the third term, 
and from the nature of the question the answer is to be less than 


$840; the fourth less than the third, therefore, the second must 
be less than the first. The second term, then, is 90 days and the 
first 9 months. Now, the first and second, although being of the 
same kind (time), are not of the same denomination; multiplying 
9 by 30 gives 270 days, and the statement is: 270 days is to 90 
days as $840 is to the required term. A glance at the two first 


terms now shows that each is divisible by 90, and consequently 


the ratio is as 3:1. So we simply divide 840 by 3, and the quo- 
tient, $280, is the answer. 
ee 840 x 90 — 75600 i8 840 x 1 — 840 
ROOF 1 280 x 270 = 75600 280 X 3 = 840 


Norse. — It may be well to remark that, in abbreviating operations in pro- 
portion by cancellation, the first and second, or the first and third terms 
(but never the second and third), are to be operated upon, 


Exam. 4. If a pole 6 feet 3 inches high casts a shadow 
74 feet long, how high is a ae whose shadow is 120 


feet long? 
fet. ft? ft. in. 
Tt : 120323 6.8 


N 240 TH 
1200 5 
100 feet. 


NN a eee 


RuLE oF THREE. 101 


In this, height is required, therefore, 6 ft. 3 in. height is put in 
the third place; and from the nature of the question, the answer 
is to be more than this third term, consequently, the second term 
is greater than the first; 120 is, therefore, put in the second place. 

Reducing the first and second terms, now, to the same denomi- 
nation, halves, and the third term to its lowest unit, inches, we 
have: 15 : 240 :: 75. 

Bringing cancellation to our aid, now, 15 and 75, the first and 
third terms, are each divided by 15, giving 1 and 5 (the 1 is 
omitted, as it does not affect the process, and the first term dis- 
appears). 

Multiplying the second and third terms, next, that is, 240 by 5, 
gives 1200, which is of the same denomination as that to which 
the third has been reduced, namely, inches. Dividing 1200 by 12 
(inches in a foot) we have 100 feet, the height of the steeple. 


240 x 75 — 18000 
1200 x 15 — 18000 


PRooF: or, or, 


120 x 64 — 750) 240 x 5 — 1200 
100 x 74 = 750 { 1200 x 1 — 1200 


The product of the means equal to the product of the extremes. 


Exam. 5. If a contractor be paid $64.35 for excavating 
148 cubic yards, or loads of earth, how much ought he to 
receive for excavating 335 loads, at the same rate? 


fé 3 
148 : 335 :: 64/35 


TO01 1005 21/45 
150}15. 

600 

75 


102 Rue OF THREE. 


The answer to this must be more than $64.35; hence, 335 isthe 
second term of the proportion. 4 

To multiply by 335, we multiply it by 8, to get 1005, a sinaple q 
multiplier. Multiplying 335 by 3, divides 64385 by 3, giving 2145, 
to be divided by 148. 

To divide by 143, it is multiplied by 7, giving 1001 for a sim- q 
ple divisor. Multiplying 148 by 7, multiplies 2145, also, by 7%, 
giving 15015, The terms, now, are; 


1001 « 1005.22. 150215 


Here, we are to multiply by 1005, and then divide by 1001. 
Now, to multiply 150.15 by 1000, we simply move the point three 
places to the right, getting 150150; and to divide this by 1000, 
we move the point three places to the left, getting for the result, 
150.150, or, as shown in the example, 150/15., the dot represent- 
ing the cipher. Hence, we simply draw the vertical line through 
the decimal point of $64.35, divide by 3, and multiply by 7, to 
get 150.15. 

Annexing a cipher (dot) to 150)15, now, we simply add the 
4-1000 (5—1), or .600; the required result is $150.75 (see Cancella- 
tion, examples 5, 6, 7 ad 8). 


Exam. 6. A builder is paid $678 for 168 perches of — 
masonry ; how much ought he to receive for 3374 perches, — 
at the same rate ? | o- 


ORD 3 
168 : 3374 :: 678 
1008 1018)" Tabpiees 
$1361/384 


In this, the answer is to be money, therefore, $678 is put in the 


RuLE oF THREE. 103 


third place; and, as more will be paid for 3374 perches than for 
168, the larger of the two is put in the middle. 

Multiplying 168 by 6, gives 1008 for a simple divisor; and 
multiplying 8874 by 8, gives 1012 for a simple multiplier. Now, 
multiplying 168 by 6, multiplies 678, also, by 6; and multiplying 
3374 by 3, divides 678 by 3. But, multiplying 678 by 6, and then 
dividing by 3, is simply multiplying it by 2. Multiplying 678, 
then, by 2, gives 1356. The analogy, now, is: 1008 : 1012 :: 1356; 
and the product of the second and third terms divided by the 
first, is the answer. 

To multiply 1356 by 1012, we annex three ciphers (dots), and 
to the result add 12 times 1356. 

To divide by 1008, we cut off the three ciphers, and subtract 8 
times 1356. 

Now, adding 12, and subtracting 8, is adding 4; so we simply 
add 4 times 1356 (in this case 4-1000), or 5424, placed in proper 
position. The multiplication by 1012 is now completed, but the 
division by 1008 is not. We have still to subtract 8 times the 
partial quotient, 5, or rather, 8-1000, which is .040. 

Subtracting 40, now, from 424, we get .384, or 38 cents; and 
adding 5 to 1356, we have $1361.38, the answer. 


Exam. 7. A drygoods dealer bought at auction 315 
yards of silk for $352.80, and at the close of the auction 
he bought, at private sale, 214 yards more of the same 
material at the same rate; how much did he pay for the 


latter 2 
a y, 
B15 : 214. 327352/80 
105 107 117/60 
235/00 
4\70 
0) 
239/70]. . 
Tht 
a 


104 RcuLE oF THREE. ihe 


The answer here is to be less than $352.80, so 214 is put for the 
second term. Dividing 315 by 8 gives 105 for simple divisor, and _ 
214 by 2 gives 107 for simple multiplier. Dividing 315 by 8 di- 
vides 352.80 by 3, also, giving 117.60; and dividing 214 by 2 


multiplies 352.80, or rather, in this case, 117.60 by 2, giving 235.20, 
The terms now are : 105 : 107 :: 235.20, and the product of the — 


second and third divided by the first gives the answer. We 
have now to add the 7-100, and then subtract 5-100; this is add- 
ing 2-100. The 2-100 of 235 is 4/70. Next, we have to subtract 
5-100 of the partial quotient, 4, to finish the division by 105; this 
is found to be .20. 

It will be observed, now, that 20 is to be added to 70, and at 
the same time 20 is to be subtracted; so we simply set down 70 
and add 235 and 4, getting 239/70. 

To find the correct decimal, or cents, now, a vertical line is 
drawn to the right of 70, and two ciphers (dots) annexed. To 
this result 2-100 is to be added, and 5-100 subtracted, as in the 
other part of the process; this is the same as subtracting 3-100. 
The 3-100 of 70 is 2110. Next, 5-100 of 2, or .10, is to be sub- 
tracted; or, simply, 2 altogether is subtracted and we get 68, the 
correct cents. The answer is $239.68. 


Exam. 8. If 3168 bushels of wheat cost $3200, what 
will 317 bushels cost at the same rate ? 


32... 
8168 : 317| :: 3200 
39 3\17 
3 


320/20 


In this, 317 bushels will cost less than 3168 bushels, therefore, 
317 is put in the second place. Then, we simply draw the vertical 
line to the right of 317, and setting that number under itself, two 


RULE oF THREE. 105 


places to the right; then 3, under the last result, two places to 
the right, and adding, we get $320.20, the required cost. 

Reason: Taking 3200 for approximate divisor, in connection 
with 3168, we find the key to the division, 1, that is, 1-100, as has 
been explained in the article on Simplified Division. Multiplying 
317, now, by 3200, and dividing the result by 3200 does not 
change 317. The divisor is not 3200, however, but 3168, so we 
must add the 32-8200, or rather its equal, 1-100 of 317; and this 
is done by simply setting 3|17, and then |08, in proper position, 
as seen in the example. 


Note.— The process will be made clear by dividing 3168, the true divisor, 
and 3200, the third term of the analogy, by 32, using the factors 4 and 8. 
It will then be seen that we have simply multiplied 317 by 100, and divided 
the product, 31700, by 99, according to Rule III of Simplified Division. 


Exam. 9. If 1372 bushels of oats cost $617.40, what 
will 1470 bushels cost at the same rate ? 


14.. 
1372 : 1470 :: 617/40 
—98 “jos 30/8700 
648|27 
12/96 
26 


661/49)00 
My 


1470 bushels will cost more than $617.40, therefore, 1470 is 
made the second term of the proportion. 

To divide by 1372, we use 1400 as approximate in connection 
with it, and the key to the division is 2 (2-100). A glance at 
1470 shows that it is divisible by 14, giving 105 for a simple mul- 
tiplier. Dividing the second term, 1470, by 14, divides the first 
term, 1372, also, by 14, or rather the approximate, 1400, giving 
100 for approximate divisor. Multiplying $617.40, now, by 105 


106 RuLE oF THREE. 


(simply set 5 times that number two places to the right, and 
add), and dividing by 100, we get 648/27. Then, adding the 
2-100 of the partial quotients, we get $661.49. Continuing the 
division, we find the correct decimal, or cents, to be .50, .98 being 
equal to 1. The answer, then, is $661.50. 


Nors.— If 1372, the real divisor, be divided by 14, instead of the approxi- 

mate, 1400 (using the factors 2 and 7 to divide by), we find the new divisor 
is 98; and that we have simply multiplied $617.40 by 105, and divided the 
product by 98. 
_ Now, since the key to the division is found by dividing the difference of 
the divisors, or 28, by 14 (the significant part of 14..), the key thus found, 
in all such cases, wéil be the difference between the approximate (new) divisor, 
found by cancellation, and the real (new) divisor, Thus, 28 + 14 = 2, the 
key; and14.. + 14= 100, the approximate (new) divisor; then, 100 —2 = 98, 
the real (new) divisor (all of which can be seen mentally). 


Exam. 10. If 2328 yards of silk cost $2460, what will 
310 yards cost at the same rate? | 


Dass 
2328 : 810]/:: 2468, 
79 17/05 
mee 
$327|56 


Notr.— The real (new) divisor in this example is 97, found mentally, thus: | 
24,. + 24 = 100, approximate (new) divisor; 72 + 24=8; then, 100—3= 97. 


Dividing the first term of the proportion by 24, divides the 
third term, 2460, also, by 24, giving 1024 for multiplier. Multi- 
plying 310, now, by 1023, and dividing by 97, we get $327.56, 
the required cost. Setting 54 times 310, or 1705, in proper posi- 
tion adds for 3 in the divisor, 97, and also for 24 in the mul- 
tiplier, 1024, and completes the multiplication. Then, 3 times 17, 
or 51, placed in proper position, is added to complete the division 
by 97. (See remarks on Rule IIT, page 38.) 


RuLE oF THREE. 107 


Comprounp PRoporrTIOoN. 


Proportion is often of such a nature as to require a 
compound and a simple ratio, and is then called Com- 
pound Proportion ; but a knowledge of simple propor 
tion will enable us to solve compound with equal facility. 

A compound ratio, as its name denotes, consists of 
two or more simple ratios, and can be reduced to a sim- 
ple ratio by the following: 


Ruiz. To reduce a compound ratio to a simple one: Multiply all 
the antecedents together for a new antecedent, and all the consequents 
together for a new consequent. 


Exam. 1. If it cost $3600 to supply 30 men with pro- 
visions for 24 days, the rations being 20 ounces per day, 
how much will it cost to supply 20 men for 36 days, the 
rations being 18 ounces per day ? 


In this example, it will be observed, there are three pairs of 
terms or couplets, namely, 80 men 


and 20 men, 24 days and 36 days, _ Noting 

men, days. 02. 
20 ounces and 18 ounces; and there 99 __ 94. 99 __ $3600 
is a single term, $3600. Arranging 90 — 36-.—18 — (2) 


the couplets, as shown in the mar- 
gin, which we will call ‘noting the question”; setting the single 
term last, and under it, the interrogation point, to indicate that 
the corresponding term is wanting, we are prepared to set the 
antecedents and consequents of the respective ratios in their proper 
places. 

The answer to the question being money, $3600 is put for the 
third term in the following statement: 


108 Rue oF THREE. 


30 : 20 :: 8600 
24: 86 
20: 18 
Tt00. : 12960 :: 3600 
4 E2540 alee 


To arrange the other terms in their proper places, now, we go 
back to the noting, and, taking the first couplet, men, in connec- 
tion with the single or third term, we proceed by question and 
answer, thus: If it cost $3600 to supply 30 men, will it cost more 


or less to supply 20 men? The answer is evidently less; 20 is - 


therefore put in the second place, and 30 in the first. Next, if it 
cost $3600 to get 24 days’ supply, will it cost more or less to get 
36 days’ supply ? More; therefore, 36 is the second term of the 
next ratio and 24 the first. Now, if it cost $3600 to supply 20 
ounces of rations, will it cost more or less to supply 18 ounces ? 
Less; and, therefore, 18 is the second term of the third ratio. 

The compound ratio now consists of three simple ones, namely: 
30 : 20, 24 : 36 and 20: 18. 

Multiplying the antecedents together, now, we get 14400 for a 
new antecedent, and the consequents together, and we get 12960 
for anew consequent. The proportion, incomplete, now stands: 


14400 : 12960 :: 83600; that is; three terms to find the fourth. — 


Multiplying the second and third terms, and dividing the product 
by the first, as in simple proportion, will give the fourth propor- 
tional, or the answer. 

Before proceeding with said multiplication and division, how- 
ever, we see by inspection that the process can be abridged by 
Cancellation, 3600 being a measure, or exact divisor, for 14400. 
Dividing each of these numbers, then, by 3600, gives 1 for the 


third term and 4 for the first, and we simply divide the second 


term, 12960, by 4, getting $3240, the required fourth proportional, 
or the answer, 


PRACTICAL PROBLEMS. 109 


Proor:  § 12960 x 3600 — 46656000 
* 714400 x 3940 — 46656000 
ue 12960 x 1 — 12960 
ae 3240 x 4 — 12960 


The product of the means equal to the product of the extremes, 
taking either pair of ratios. 


Nore. — It is almost needless to remark, at this stage of the work, that, 
before multiplying the antecedents and consequents together, as in the 
example, recourse may be had to cancellation, dividing the first and second 
terms, or the first and third, by any numbers that wili reduce or eliminate 
those terms. 


PRACTICAL PROBLEMS. 


Proportion may be greatly abridged by the use of 
aliquot parts. P 


An Aliquot Part of a quantity is such a part as, when taken a 
certain number of times, will exactly make that quantity. Thus, 
4 is an aliquot part of 12, 24 of 10, 123 and 334 of 100, etc. 


Exam. 1. What is the cost of a cargo of iron weighing 
259166 pounds, at $36.16 per gross ton? 


There are 2240 pounds in a gross ton, and by proportion, the 
statement is: 


2240 : 259166 :: 36.16 


Multiplying the second and third terms, and dividing by the 
first will give the solution, which, by the usual method, is quite 
tedious. 

Now, by assuming as the price per ton, some measure, or 


hee 
110 PRACTICAL PROBLEMS. } 


aliquot part, of 2240, as, for instance, 82, the foregoing problem 
can be abridged, thus: 


70 i 
"99-40: 2591616 3: 3 
3702/3871 — price @ $32 00 
462'796 — “©  & 4.00 
11/569-—= * as 10 
5/785 a Sees O05 
INS ae 01 


$4183/688 — “ “ $36 16 


Tf the price per gross ton were $32, the statement would be: 
2240 : 259166 :: 32, and by cancellation, this becomes — 

70 : 259166 :: 1; 82 being contained in itself once, 
and 70 times in 2240. Monae by 1, now, and dividing by 
70, gives the price of the given number of pounds at $32 per ton. 
To divide by 70, we cut off the cipher, and also one figure, 6, 
from the right of the dividend (this divides by 10), then dividing 
by'7, we get $3702.37, the price at $32 per ton. Next, we take 
aliquot parts of this for $4.16, the difference between the real and 
the assumed price, thus: $4 is 4 of $32, therefore, the price at $4 
will be the eighth part of that at $32, or, $462.79+ (the remain- 
ing decimals being mills, etc.). Then, 10 cts. is 7, of $4, 5 cts. is 
4 of 10, and 1c. is4of 5. The sum of the several pare quo- 
tients, or, $4183.68, is the price at $36.16. 


Exam. 2. What is the cost of a cargo of iron weighing 
276388 pounds at $23 per gross ton ? 


2'7638|8 . 
8948/40 = price @ $32 
~O8710 PCOS 66 $8 
128)38875 = “ “ J 


POS8791. = «. * B98 


PRACTICAL PROBLEMS, 111 


Here, we simply divide the given number of pounds by 70, as 
in the preceding example, and we have the price at $32 per ton. 
The difference between this and the real price, $23, 1s $9, for 
which we take aliquot parts, thus: $8 is + of 32, and $1 1s 4 of 
$8. Adding the price at $8 and $1, gives the price at $9, which, 
being taken from the price at $32, gives that at $23, or $2837.91. 


Norse. — The subtraction of the two-numbers is performed at a single 
operation, as pointed out in short methods for Multiplication. (See note 
to example 14, page 78.) 


Exam. 3. What is the cost of 24840 pounds of coal at 
$3.50 per gross ton ? 


The statement of this problem by Proportion would be: 


9240 : 24840 :: 3.50 | 


and the product of the second and third terms, divided by the 
first, would be the answer. 

We will here assume $28 as the ie per gross ton, and we 
have: 


8.0 ii 
9340 : 2484/0 :: 28 
810150 — price @ $28 
pes UR aC 1 


' By cancellation, 2240 and 28 become 80 and 1; 28 being con- 
tained in itself once and 80 times in 2240. Dividing then by 80, 
as in the two preceding examples, we get $310.50, the price at 
$28 per ton. Now, $3.50 is 4 of 28; therefore, the eighth part 
of the price at $28 is the price at $3.50. Dividing $310.50 by 8, 
we get $38.81, the required cost. 


Nore. —It may be well to remark, in connection with this method of ab- 
breviation, that, although there are other aliquot parts, or measures, of 
2240, the nnmbers 32 and 28 will be found the most convenient for practical 


PB yy PRACTICAL PROBLEMS. 


purposes. The price per gross ton will always decide which of the two 
the more desirable to be taken as the assumed price. 


FREIGHT. 


When the price per gross ton is less than a dollar, we 
would proceed as in the following: 


Exam. 4. What is the freight on 23800 pounds of mer- 
chandise at 96 cents per gross ton ? 


23/8010 
8/40 = price @ 32 ©. 
$10/200— * + * 96e., 


If the price per ton were $96, instead of 96 cents, in this 
example, we would cut off the right hand cipher, and divide by 
7, as in the other examples, getting the price at $32. But cents 
are the hundredths of dollars, so we cut off two more figures from 
the right, and then proceed as in the other examples. In all such 
cases, then, three figures are cut off from the right, and the process 
will be as pointed out in the foregoing examples. $3.40 is the 
freight at 32 cents, and 3 times this, or $10.20, is the freight at 
96 cents. 


Oats. 


What is the cost of 24160 pounds of oats at 58 cents 
per bushel, the bushel consisting of 32 pounds? 


This problem, by Proportion, would be: 


é 


32 : 24160 :: 58 


and the product of the second and third terms, divided by the 
first, would be the answer. 


PRACTICAL PROBLEMS. * Lie 


Assuming that the price per bushel is 382 cents, instead of 58, 


the statement would be: 

32 : 24160 :: 82; and this by cancellation would be: 
1; 24160 :: 1. The price of the given number of 

pounds, then, at 32 cents per bushel is 24160 cents, or $241.60; 

that is, 1 cent per pound, and the process thus: 


241/60 — price @ ae 
433/20 = “ 
420 90 66 66 
15}10 
437190 — “ & 


| 


lewl® 


At 1 cent per pound, or 82 cents per bushel, the price is $241.60. 
Doubling this gives the price at 64 cents per bushel, which is 6 
cents per bushel too much. Then 4 cents is 4 of 82; dividing 
$241.60 by 8 gives the price at 4 cents. Next, 2 cents is } of 4; 
taking 4 of $30.20, the price at 4 cents, gives $15.10, the price 
at 2 cents, and deducting both from $483.20, the price at 64 cents, 
gives $437.90, the price at 58 cents, or the answer. Or thus: 


241/60 32 
120/80 16 
60/40 8 
15|10 2 
437/90 58 


The process here is self-evident and requires no analysis. 


Note. — The four last problems will suggest the method of abbreviating 
commercial calculations by the use of aliquot parts. Wheat, buckwheat, 
barley, etc., may be treated in like manner, 


PERCENTAGE, 


Per cent. from the Latin per centum, signifies by the 
100. In business transactions, it means a certain part of 
every 100. Thus, 2 per cent. means 2 of every 100, and 
may signify 2 cents of every 100 cents, 2 dollars of every 
100 dollars, 2 yards of every 100 yards, ete. 

The character, %, is used in business transactions to 
represent the words per cent.; thus 2% means 2 per cent. 

In Percentage, five quantities are concerned, namely : 

The Base, Rate per cent., Percentage, Amount, and 
Difference. 

The Base is the number on which the percentage is 
computed. 

The Rate per cent. is the part of 100 taken. 

The Percentage is the fourth proportional to 100, the 
rate per cent., and the base, taken in the order mentioned, 

The Amount is the base plus the percentage. 

The Difference is the base less the percentage. 


Case I. 
Given, the base and rate, to find the percentage. 


Exam. 1. What is 4% of $350. 
350 x 4 
$1400 


PERCENTAGE. 115 


To solve this, we multiply the base, $350, by 4, the rate per 
cent., and divide the product, 1400, by 100; simply cutting off 
two figures from the right; $14 is the percentage. 

Reason: The foregoing is simply a question in proportion, ex- 
pressed thus: If $4 be allowed on $100, how much ought to be 
allowed on $350, at the same rate? 

And the analogy is: as $100, base, is to $350, base, so is $4, the 
percentage on $100, to the corresponding percentage on $350. 

Thus: 

100 : 850 3: 4 


er, as $100 is to its percentage, 4, so is $350 to its percentage. 
Thus: 
100 : 4:3: 3850 


In either case, the product of the second and third terms 
divided by the first gives the fourth proportional. Hence, 

Rue. To jind the percentage: Multiply the base by the rate per 
cent. and divide the product by 100. 


Exam. 2. A.’s salary is $2500 a year; if he spend 10% 
for board, 6% for clothing, 5% for books, and 9% for other 
purposes, what are his yearly expenses ? 


-Nore.— When several rates refer to the same base, they may be added or 
subtracted, according to the nature of the question. Thus: 10% + 6% + 5% 
+- 9% = 80%; then 30% of $2500 equals $750, his yearly expenses. 


Case II. 
Given, the percentage and base, to find the rate. 
Exam. 1. What per cent of $350 is $14? 


The question, fully expressed, is this: If $14 be allowed on 
$350, how much ought to be allowed on $100 at the same rate ? 


116 PERCENTAGE. 


And the analogy is: As $350, base, is to $100, base, so is $14, 


the percentage on $350, to the corresponding percentage on $100. 
Or, as the given base is to its percentage, so is $100, considered 
as a base, to its percentage. Thus: 


350 : 14 :: 100 


Multiplying the second and third terms, now, and dividing by 
the first, we get the rate on $100, or the rate per cent. Hence the 

Rue. Multiply the percentage by 100 and divide the product by 
the given base ; the quotient is the rate. Thus: 


1400 + 350 — 4% 


In the example, the percentage is 14 and the base 350. Annex- 
ing two ciphers to 14; that is, multiplying i¥ by 100, the third 
term of the analogy, we have 1400; and dividing this by 350 gives 
4, the rate per cent. 


Exam. 2. A merchant, failing, owes $14300, and his 
assets are only $10725; how much on the dollar can he 
pay to his creditors ¢ 


_ 143/00 107 25] vs 
1001 T5|UT5 
[ts 
In this, $10725 is the percentage, and $14300 the base. An- 
nexing two ciphers (dots) to the percentage, we have 1072500 to 
be divided by 14300. Cutting off the ciphers from both numbers 
and multiplying the remaining parts each by 7, we have 75075 to 
be divided by 1001. 


The quotient is 75%; that is, 75 cents on the 100 cents, or the 
dollar. 


PERCENTAGE, 117 


Case III. 
Given, the percentage and the rate, to find the base. 


Exam. 1. $14 is 4% of what number ¢ 


The question, fully expressed, is this: If $4 be allowed on 
$100, on what sum ought $14 to be allowed? And the analogy 
is: As $4, the percentage on $100, is to its base, 100, so is $14, 
the given percentage, to its corresponding base. Thus: 


4:100:: 14 
Hence the 
Rue. Multiply the percentage by 100 and divide the product by 
the given rate ; the quotient is the base. Thus: 


1400 + 4 — $350, the base. 


Casse LY. 


Given, the amount and rate, to find the base. 


Exam. 1. What number increased by 4% of itself is 
equal to 364 4 


ANALYsIS: Since the amount is the base plus the percentage, it 
is evident that 104 is the amount of 100 (considered as a base). 
The question now is: If 104 be the amount of 100, of what num- 
ber is 364 the amount ? ~ 

And the analogy is: As 104, the amount of 100, is to its base, 
100, so is 864, the given amount, to its corresponding base. Thus: 


104: 100 :: 364 


Multiplying the second and third terms now, and dividing the 
product by the first, will give the base. Hence the 


118 PERCENTAGE. 


RuiE. Multiply the given amount by 100 and divide the product 
by 100 plus the rate; the quotient is the base. ‘Thus: 


364|.. + 104 
14/56 
349/44 
60 
350\ 04 
04 


Annexing two ciphers (dots) to 364, we have 36400 to be 
divided by 104. The quotient is 350, the required base (Rule II, 
Division). 


Exam. 2. A merchant increased his stock in trade ‘by - 
127% of itself, and then had $3800; how much had he at 
first ? , 

BSHO. 2-54 ULZ 
3420|000 O00 + 1008 
27/360 
$3392/640 
224 
864 


In this, $3800 is the amount, and 12 the rate. Annexing two 
ciphers (dots) to 3800, we have 380000 to be divided by 112. 
Multiplying both by 9, the divisor becomes 1008. The answer is 
$3392.86. 


Case VY. , 
Given, the difference and the rate, to find the base. 


Exam. 1. What number diminished by 4% of itself, is 
equal to 336 ? 


PERCENTAGE, 1192 


ANALysIs: Since the difference is’the base less the percentage, 
it is evident that the difference of 100 (considered as a base) is 96. 
The question, now, is: If 100 be the base and 96 the difference, 
what ought to be the proportional base for the difference, 336? 

And the analogy is: As the difference, 96, is to its corre- 
sponding base, 100, so is the given’ difference, 336, to its corre- 
sponding base. Thus: 


96 : 100 :: 336 
Multiplying the second and third terms now, and dividing the 
product by the first, will give the base. Hence the 
Rue. Multiply the given difference by 100 and divide the product 
by 100 less the rate; the quotient is the base. Thus: 


336]... + 96 
13\44 
52 
349/96 

Annexing two ciphers (dots) to 336, we have 33600, to be di- 
vided by 96. The quotient is 3493%, or rather 350, the remainder 
being 1, 

APPLICATIONS OF PERCENTAGE. 

The five rules of Percentage now established can be 
readily applied to problems in Commission, Stocks, Profit 
and Loss, Taxes, Insurance, etc. A few examples will 
suflice. 


1. A merchant in Albany remits to his agent in 
Chicago $850.75 for the purchase of grain. The remit- 
tance includes commission at 24%; how much will the 
agent expend for grain, and what will be his commission 4 


In this, $850.75 is the amount, and 24 the rate. Now, what is 
the base ? 


erry: 
os 
aoe 


120 PERCENTAGE. 


To this, Case IV is applicable. Thus: 
850|75 + 1025 
21/25 

29/50 
525 
025 


Anatysis: In this, we have to multiply first by 100, and then 
divide by 100 (in 1024), consequently, $850.75 undergoes no 
change and we simply draw the vertical line through the decimal 
point and make the proper correction for 24. To multiply 850 by 
24, we conceive a cipher annexed and divide by 4 (4° — 24), or, 75 
of 850 as it stands, gives 21.25, which is placed in proper position 
and subtracted. Then, 7/5 of 21, or .525, is placed in position 
and added, and finally, the 71, of 1 (carried in adding |50 and 
525), or |025, is subtracted. The result is $830, the base of com- 
mission, which is the sum to be expended for grain. Then, 
24% of $830 — $20.75, or $850.75 — $830 — $20.75, the com- 
mission. . 

Note.— Commission is charged only on what is expended or collected by a 
person acting in the capacity of agent. 

2. An agent sold real estate on commission at 3%, and 
returned to the owner, as net proceeds, $2425; what was 
the price received for the property, and what was the 
commission ? 

In this, the net proceeds, $2425, is the difference and 8 the rate. 
Now, what is the base ? 

Here, Case V is applicable.* Thus: 

2425|.. + OF 
72|75 
2)16 
6 
$2499197 


PERCENTAGE. 121 


Annexing two ciphers (dots) to 2425, we have 242500, to be 
divided by 97. The quotient, by simplified division, is $2500 
(the remainder, .97, being 1), the required base, or what the prop- 
erty sold for; whence, by subtraction, we obtain the commission, 
$75. Or, 38% of 2500 = $75. 


Note.— From a due consideration of the foregoing article, the student 
cannot fail to appreciate the great advantage of a thorough knowledge of the 
principles of proportion (given under the head of ‘‘The Rule of Three” in 
this work) én all cases where percentage is concerned ; for, if a set rule should 
be forgotten, as frequently happens, a knowledge of proportion will enable 
us to recall it without difficulty. 

If, for instance, we should forget the set rule for the last problem, we 
would reason thus: If, from property which sold for $100, I receive $97 as 
the net proceeds, from what sum ought I to receive $2425 as net proceeds? 

And the analogy is: 97: 100:: 2425, which gives the rule at once. 


EXAMPLES FOR PRACTICE. 


1. What is 123% of $5600? Ans. $700. 
2. What per cent. of $720 is $21.60? Ans. De 
8. 18 is 25% of what number ? Ans. 12. 


4, What number increased by 15% of itself is equal to 
644 ? Ans. 560. 


5. What number diminished by 10% of itself is equal 
to 504% Ans. 560. 


Notr.— By solving those five examples by Proportion, the reason of the 
set rules will be impressed upon the memory. 


INTEREST, 


In computations in Interest there are five quantities to 
be considered, namely: the Principal, the Interest, the 
Amount, the fate and the Zime. 

Principal has reference to money (or its equivalent) 
lent by one person to another, on condition that the bor- 
rower pays a certain sum to the lender for the use of the 
money. 

[Interest is the sum paid for the use of the principal, — 
and is calculated on the basis of $100 as a standard prin- 
cipal, and one year as the time. 

Amount is the principal and interest together. . 

Rate, or Rate per cent. per annum, is the sum allowed 
for the use of $100, for a year, per cent. meaning by the 
100, and per annum, by the year. 

The Time is that agreed upon by the parties to the 
transaction. 


The most important problem in computations in interest is that 
in which the principal, the time and the rate are given, to find the 
interest or amount. 


Exam. 1. What is the interest of $376 for 2 months 
at 6 per cent 2 | 


INTEREST. 123 


ANALYsIs: This, in substance, is a question in compound pro- 
portion, and when fully expressed reads thus: 

If $6 be paid for the use of $100 for 12 months, how much 
ought to be paid for the use of $376 for 2 months, at the same 
rate ? 

The statement, by proportion, would then be: 


THOS 316-226 
Te 2 


A moment’s glance at the terms of this proportion shows that 
the process can be abridged by cancellation; thus, dividing 12 
and 6 each by 6 they become 2 and 1; next, dividing 2 thus found, 
and 2, the second term of the second ratio, each by 2, they dis- 
appear, and the proportion then is simply: 


TOs 30 10s 


Multiplying the second and third terms, in this proportion, and 
dividing the product by the first, we get $3.76, the fourth pro- 
portional, or the interest for 2 months. 


From these facts we see that, when the rate per cent is 
6, and the time 2 months, or 60 days (30 days to a month), 
the interest of any principal will be just as many cents 
as there are dollars in the gwen principal. 


Taking this for our basis, it is evident that the interest for any 
number of months, at 6%, will be the interest for two months 
multiplied by half the given number of months, thus: 

The interest of $376 for 10 months, at 6%, is $3.76 multiplied 
by 5 (half of 10 months), or $18.80; in other words, the interest 
for 2 months multiplied by 5 will be the interest for 10 months. 

The interest of $376 for 3 years and 4 months, at 67, is $3.76 
multiplied by 20 (half of 40 months, the number in 3 years and 
4 months), or $75.20. 


124 INTEREST. 


And the interest of $376 for 3 years, 4 months.and 15 days, at 
6%, is the interest for 3 years and 4 months plus + of $3.76 (the in- 
terest for 60 days), 15 days being } of 60, thus: 


_$3/76 — the int. of $3876 for 2 mos., or 60 days, at 6%. 


75|20 — - 3 years, 4 mos., or 40 mos, 
yy ee ““ «“ 15 days. 
$76|14 — e : 3 years, 4 mos., 15 days. 


From this the interest at any other rate per cent. can be readily 
obtained by the method of aliquot parts; thus, if the rate were 
7%, add 4 of the interest at 67, and if at 5%, deduct the 4; if at 
8%, add 4 (2 being + of 6), and if at 4%, deduct 4; if 9%, add 4, 
and if 3%, take the 4; 74%, add 4 (14 being + of 6), and if 442, 
deduct the 4, ete. 


Nore. —It need scarcely be remarked that, in computing interest, the 
business method is to reject the cents of the principal when less than 50, 
and when 50 or more, add $1. 


Exam. 2. What is the interest of $239.97 for 1 year, 5 
months and 17 days @ 6%? 


$2|40 — the interest for 2 mos., or 60 days. 


20/40 = = “1 year, 5 mos., or, 17 moa: 
60 = “ 6 15 days. 
08 == % e 2 days.. 
$21/08 — . 1 year, 5 mos., 17 days. 


Here, we call the principal $240; the interest on this for 2 
months, or 60 days, is $2.40. Multiplying this by 8} (half of 17 
months, or 1 year and 5 months) gives $20.40. For 17 days we 
take aliquot parts of 60, thus: 15 days = 4 of 60; the $ of $2.40 
is 60 cents; then 2 days — 4), of 60; the {; of $2.40 is 8 cents, 


INTEREST. 125 


or .08; and by addition we get $21.08, the interest for 1 year, 5 
months and 17 days. 
Nott. — Should it be desirable to retain the cents of the principal, it will 


not, of course, affect the process, the figures to the right of the line being 
simply decimals, or cents and mills. 


From the foregoing principles and examples we derive 
the following 


Easy Meruop or Computine InTEREst oN ANY SuM, FOR 
ANY Timz, AT ANY RATE PER CENT., oN A Basis oF 360 
Days To THE YEAR. 


GENERAL RuLeE. (1.) Draw a vertical line through the given prin- 
cipal, two places to the left of units; the result is the interest for 2 
months, or 60 days, at 6%: (2.) multiply this by half the months in 
the given time (reducing the years, if any, to months): (8.) add, for 
days, such parts of 60 days’ interest, as the days are aliquot parts of 
60; the result will be the interest for the given time at 64%, the dollars 
being to the left, and the cents and mills to the right of the vertical 
line. 

Then, for any rate other than 6%, add or subtract the proper 
proportions, as pointed out in the foregoing analysis. 


Nore.— If the number of days be not an aliquot part of 60, say 25 days, 
for instance, then say 20 days — ¢ of 60, and5— ¥f of 20; the two results, 
when added, make 25. If 27 days, then 30 = 1g of 60, and 3 = qs of 30; 
the difference is 27 (30 — 3), ete. 


The foregoing is also the rule generally used for Bank or Busts 
ness Discount, as it is called. 


Exam, 3. What is the bank discount on a note drawn 
at 3 months for $480.23 at 64 ? 


126 INTEREST. 


4|S0 = the interest for 2 mos,, or 60 days. 


2)40 == et eT m0., OF 30 ave. : 
2 FS bs 8 days (grace). 
$7|44 = . i 93 days 


Deducting the interest (discount), $7.44, from $480.23, gives 
the present worth or the net proceeds. 


Exam. 4. What is the interest of $468 for 1388 days 
at 6% % 


4168 
9136 13/8 + 3]0 
12d — } of 60 93 “Als 
6d = 4 of 12 468 
$ 10/76 


Dividing 138 days by 80, as shown in the margin, we get 4 
months and 18 days, the process is then according to the rule. 


Now, since the interest of any sum of money for 60 days, or 
2 months, at 6%, is as many cents as there are dollars; that is 1%. of 
the principal, the interest for 6 days is a tenth of that for 60 days, 
or 2 months, and for 1 day, a sixth of that for 6 days; for 600 days, 
or 20 months, the interest is ten times that for 60 days, or 2 months 
and for 6000 days, or 200 months it is ten times that for 600 days, or 
20 months; in other words, any sum of money will double itself, or 
the interest will equal the given principal, in 6000 days, 200 months, 
or 162 years, at 6%, simple interest, thus: 

The interest of $5480 for 6000 days or 200 months = $5480. 
ck ‘* $5480 for 600 days or 20 months = $548.0 
ee « $5480 for 60 days or 2 months = $54.80 
es é«° $9480¢ for Go days! s.06 ts eee = $5.480 
a ‘+ 5480 for 4 days. ieee eee et Oe 

And from this the interest for any time can be readily found, as 
illustrated in the following: 


: 
4 
q 
; 


INTEREST. 


127 


Exam. What is the interest of $5498.70, for 3 months 


and 10 days, at 6 per cent.? 


In this, counting 30 days to the month, the 
time is 100 days, and we simply cut off one 
figure from the dollars to get the interest for 
600 days; then a sixth of this, or $91.645, is 
’ the interest for 100 days, or 3 mo., 10 da. 


$049|8 . 70—600da. 
$91} 645 —100 ** 


Exam. What is the interest on a note for $7560, dated 
Jan. 31, 1906, and payable May 5th, following, at 6 per 


cent.? 


Counting Feb, 28 days, Mar. 31, April 30 and 
May 5, the time is 94 days; and the interest for 
100 days is $126 from which $7.56, the interest 
for 6 days is deducted; this leaves $118.44, the 


interest for 94 days. 
Or, 


The interest for 60 days=$75.60; for 30 days 
$37.80; for 3 da. a tenth of this, or $3.78 and for 
i da. a third of this, or $1.26, making, when 


added, the required interest. 


$756/0 —600da. 


126) =-100 © 
Too Ons 


$11s|44— 94 °° 


$75|50=60 da. 
37/80=30 << 
SSS oo 

1 26= 1 cé 
$118 44—94 da 


Exam. What is the interest of $5840, for one year, 8 


months and 15 days, at 7 per cent.? 


In this, 1 yr., 8 mo. = 20 mo. and at 6%, 
the interest is $584. 

Then, the interest for 15 da. at 6% is one 
fourth of 60 days’ interest, or of $58.40; this 


gives $14.60 which is added, making $598.60. 


for the given time at 6% To this we add the 
interest at 1%,or one-sixth of that at 67, $99.766, 
which gives $698.366, the interest at 77. 


$584|0 =20 mo. 


14/60 =15 da. 
$598|/60 —67_ 
y9|766=17 


$6981366 =77 


Nots. —If the rate were 5%, $99.766, or one-sixth of 6% would be deducted; 
if 4%, a third of 6% or 2% would be deducted; if 744% a quarter of that at 6% 
would be added 144 being a quarter of 6, etc. (See page 124.) 


128 INTEREST. 


RuLE. To compute interest at any rate per cent. for any number 
of days, on the basis of 360 days: Multiply the principal by double 
the rate; multiply the result by the number of days; cut off fiwe 
jigures from the right, counting from the decimal point always, and 
add a third and a siath of that third. 


Exam. What is the interest of $50000 for 14 days, at 
34 per cent.? 


In this, multiplying by 7 (double 33) and then by 
14, is the same as multiplying by 98 (7 x 14) at once. $50000 x 98 
Multiplying 98 by 5, and annexing the ciphers, we 49100000 


get $4900000. Cutting off five places, and adding s pee 


a third and a sixth of that third, we get the required . $68|05500 
interest, to five places of decimals, 


And to compute interest at any rate, for any number of days, on 
the basis of 365 days, in other words, exact interest : 

Ruue. Multiply the principal by double the rate; multiply the result 
by the number of days; cut off five figures from the right, counting 
from the decimal point always, and add a third, a tenth of that third 
and a tenth of that tenth. 


Exam. What is the interest of $256800 for 1 day at 
2 per cent. (865 da.) ? 


Here, the time being only 1 day, we simply multi- $256800 x 4 
ply by 4 (double the rate); then, cutting off five ee pe 
places, and adding a third, a tenth of that third and aie 


1384240 
a tenth of that tenth, we get the interest, correct to 3494 
five places of decimals. $14/07264 


Norsk. — This rule will be found useful in finding the interest on daily 
balances. (See page 220.) 

Both the foregoing rules will be found more fully explained in the Appendix 
(pp. 177 and 178) together with the reason of the rules, 


PaRTIAL PAYMENTS. . 129 


PARTIAL PAYMENTS OR INDORSEMENTS. 


The United States Courts have decided that, 

I. “The rule for casting interest when partial pay- 
ments have been made, is to apply the payment, in the 
first place, to the discharge of the interest then due. 

Il. “If the payment exceeds the interest the surplus 
goes toward discharging the principal, and the subsequent 
interest is to be computed on the balance of the principal 
remaining due. 

III. “If the payment be less than the interest the sur 
plus of interest must not be taken to augment the princi- 
pal, but the interest continues on the former principal until 
the period when the payments, taken together, exceed 
the interest due, and then the surplus is to be applied 
toward discharging the principal, and the interest is to be 
computed on the balance as aforesaid.” — Decision of 
Chancellor Kent. 


EXAM. 


$1000. Arnany, N. Y., May 1, 1885. 


Two years after date I promise to pay to A. B., or order, one 


thousand dollars, with interest, value received. 
Cru: 


On this note were indorsed the following payments: 


Jan. 1, 1886, received $150. 
Sept. 1, 1887, received 5. 
Jan. 1, 1888, received 200. 


130 PARTIAL PAYMENTS. 


What remained due May 1, 1888? 


Process : | 
Principal on interest May 1, 1885........ ..... ee $1000 
Interest from May 1, 1885, to Jan. 1, 1886 (8 mos.) @6 2, 40 
Amount ee $1040 
First. payment, Jan. 1) 18860-54000 2 aes vin ghee oes 150 
New principal...... $890 
Interest from first payment to Sept. 1, 1887 (20 mos.).. 89 


Second payment, Sept. 1, 1887 (less than interest). $75 
Interest on $890 from Sept. 1, 1887, to Jan 1, 1888 (4 mos.) 17 80 


Amount ...... staea- SOGGH ER 
Third payment.January 1, 1888°.5..4 4-2 $200 
Sum of second and third payments ........ Spi oS 275 
New principal...... $721 80 
Interest from Jan. 1, 1888, to May 1, 1888 (4 mos.)... 14 44 
Balance due May 1; 1888 . 2. °.. 5520 Ses eee $736 24 
Hence the 


Rue I. Compute the interest on the given principal to the time of 
the first payment, and, if less than the payment, add it to the prin- 
cipal and subtract the payment from the amount, the difference will 
be the new principal. 

Il. But if the payment be less than the interest, let the account 
stand (noting the payment on the document) tal the next payment, 
when, if the sum of the payments shall equal or exceed the interest then 
due, add the interest to the new principal and subtract the sum of the 
payments from the amount; the difference will be the new principal, 
with which proceed as before. 


Nots,— This rule applies to bonds, mortgages and other obligations bear- 
ing interest. 


BANK DISCOUNT. 


Bank Discount is an allowance made to a bank for the 
payment of a note before it becomes due. 

The money received for the note when discounted is 
the Proceeds, and is equal to the face of the note less the 
discount. 

The Face of a note is the sum made payable by the 
note. 

Three days, called Days of Grace, are allowed on a 
note, after the time it is nominally due, before it is 
legally due. 

Thus, a note drawn on April 8, at 2 months, would 

not be legally due till June 11. 
~The person by whom a note is signed is the Maker or 
Drawer of the note. 

The person in whose favor, or to whose order the note 
is payable, is called the Payee; and the folder is the 
owner of the note. 

The Maturity of a note is the expiration of the days 
of grace. 

To J/ndorse a note is to have the payee or holder write 
his name across the back of it. 


182 BANK DISCOUNT. 


Note. — An indorsement makes the indorser liable for the payment of a 
note ifthe maker fails to pay it when due. 


The following is the usual custom of borrowing money at banks: 
The borrower presents a note, either made or indorsed by himself, 
payable at a future specified time. Interest is calculated on the 
face of the note for the time it has to run from the date of discount , 
this interest is deducted from the face and withheld by the bank 
in consideration of advancing the money before the note matures. 
Hence the 

Rue. To find the discount and the proceeds of a note: 

I, Find the interest on the face ofthe note for three days more 
than the time specified ; this interest is the discount. 

II. Subtract the discount from the face of the note; the differ- 
ence is the proceeds, 


Notr. — When a note is given in settlement of an account, the business 
method isto add the interest to the debt and draw the note for the full amount. 
But sometimes notes are drawn promising to pay ‘‘ with interest’ (not in- 
cluding the interest with the debt). In such cases the amount, that is, the 
debt and interest together, is the face of the note, or sum made payable, 
and must be made the basis of discount. 


EXAM. 
— $1260.15 AuBany, N. Y., Jan. 17, 1889. 


Ninety days after date I promise to pay to the order of Case & 
King, one thousand two hundred and sixty and ;4%, dollars, at 
the Exchange Bank, for value received. 

JOHN JONES. 

Suppose Case & King discounted the foregoing note on Febru- 
ary 4, what would be the discount, and what the proceeds ? 


Anauysis: In the first place we find the day of maturity, thus: 
Counting 90 days after January 17, we find the nominal day to be 
April 17, thus: subtracting 17 from 31 leaves 14 days for January, 
then February 28 days, March 31 days, and 17 days of April make 


BANK DISCOUNT. 133 


90. To this we add 3 days grace, and the day of maturity is 
April 20. Interest is computed on the note, now, from the date 
of discount, February 4 to April 20, at 6%. The number of days 
from February 4 to April 20 is 75, and the interest of $1260.15 
for 75 days at 6% is found thus: 


12|60 — interest for 60 days. 
3|15 = A t Daas 


$1575 Tg’ 


The discount, then, is $15.75, and the proceeds $1260.15 minus 
$15.75, or $1244.40. 


Norse. —For business method of computing interest, see note, page 124; 
also general rule for interest, page 125. 


Although the foregoing rule is that which is employed 
in actual practice, it is founded on a principle radically 
false, and always gives the discount too large, and conse- 
quently the present egret too small, °Y the interest of the 
true discount. 

The true present worth of any debt is such a sum as 
would, if lent at interest at the assigned rate, amount to 
that debt at the time at which it would have been due. 
Hence the 


Rute. To find the true present worth of a debt: As $100 plus its 
interest for the given time, and at the given rate per cent., is to 
$100, so is the debt to its present worth. 

Thus, the true present worth of the note in the last example is 
found as follows: 

The interest of $100 for 75 days at 6% is $1.25; adding this to 
$100, and proceeding according to the rule, we have the following 
analogy: $101.25 : $100 :: $1260.15 : $1244.59, the present worth. 


134 BANK DISCOUNT. 


Subtracting $1244.59 from $1260.15, the face of the note, gives 
$15.56, the true discount, or 19 cents less than was found by the 
bank method. And the interest of $15.56, the true discount, for 
75 days at 6%, is found to be 19 cents. 

The reason of this rule: $101.25 is the amount of $100 for the 
given time and rate, and $1260.15 is the wmount of $1244.59 for 
the same time and rate; that is, the amount of $100 is to the prin- 
cipal which produced that amount as any other given amount is to the 
principal corresponding to the said amount. 

It is evident, also, that we might use the amount of $1, and $1 
itself, or the amount of any sum whatever, and that sum itself, for 
the two first terms of the analogy; but it is generally more easy 
to use $100 and its amount. : 


Rute. To jind the face of a note, the proceeds being given: As 
the proceeds of $100 is to $100, so is the proceeds of the note to 
its face. 


Exam. 1. For what sum must a’note be drawn at 4 
months to net $750 when discounted at 6% ? 


Norz.—The days of grace having been abolished in the State of New 
York, are not taken into account in the following examples. 

ANALYSIS: The interest of $100 for 4 mo. at 6% = $2. 
Deducting this from $100 gives $98, the proceeds of $100. 


Then, As $98: $100: :$750: x ="“X"™ _ 9765.30 
the sum for which the note must be are : 


that is, the proceeds of a note whose face is $100, ts to that face, as the 
proceeds of any other note ts to its face.. 


The division by 98 is performed by the 750/00 — 
simplified process : Dividing first by 100, 15/00 
2 times 750 = 1500, and 2 times 15 = 30; are set 30 


in proper positition and added; the sum is $765.30 —- 
$765 30 


(For a more simple method to find the face of a note, see p. 214.) 


Banx Discount. 135 


RULE.—To find the interest corresponding to a given rate of bank 
discount : 


I. Assume $100 as the amount or face of the note, and find the 
discount and the proceeds of that amount for the time the note has 
torun. Then, 

II. Apply the following proportion : 

As the given time : 1 year : 
the principal : $100 the interest ; the rate. 


Exam. What rate of interest is paid, when a note 
payable in 60 days is discounted at 2% a month ? 


ANALYSIS: 2% per mo. = 24% per year. 

The discount of $100 for 60 da., or 2 mo., at o4y = $4. 
Deducting this from $100 gives 896, the ihenede! 

Here, then, having assumed $100 as the amount; $96 is the prin- 
cipal, $4 the interest and 2 mo. the given time, and we have the 
following proportion : 

As 2 mo. : aC aeeat : 
$96 : $100 ; 

Reducing this to a simple proportion, we have: 192 :1200:: 4 
Multiplying the second and third terms, now, and dividing by the 


first, we have: “1” 4 = 4800 + 192 = 25¢, the rate. 


Or, having found! the discount and the proceeds of $100; draw a 
vertical line and set the given time and the proceeds on the left; 


: $4 : the rate. 


and on the right sct 1 year, $100 and 2 mo.|12 mo 
the discount. Divide the product of: $96) $100 

the numbers on the right by the product $4 

of those on the left, and the quotient is 192/4800 (25% 
the rate. 


NotTe.—The reader need scarcely be told that the numbers on both sides 
may be cancelled, if desirable; in this case they can be cut down to 
100 + 4 = 25%. 


136 Banx Discount. 


ID Pag Mal 
‘ rt. 
a 
AN 
4 
| b, 
ay 


RvuLE.—To find the rate of bank discount corresponding to a given 
rate of interest: 


Assume $100 as the proceeds of the note and find the interest and 
the amount of that for the time the note has torun. Then, . 


As the pe ly at : : the interest : the rate. 
the principal : $100 


Exam. A broker buys 30 day notes at such a discount 
that his money earns him 24% a month; what is the rate 
of discount ? 


ANALYSIS: 24% per mo. = 30% per year, 
The interest of $100 for 30 da. or 1 mo. at 30% = $2.50. 
Adding this to $100 gives $102.50, the amount, 

Here, then, $102.50 is the principal, $2.50 the interest and 1 mo. 
the given time, and we have the following proportion: 


As 1 mo, : 12 mo. os $2 50 : the rate. 
$102.50 : $100 


and this now becomes: $102.50 : 1200 : ; $2.50 == 700*250_soq000 
+ 10250 = 29147, the rate. 


Or, having found the interest and the amount of $100; draw a _ 
vertical line and set the given time and the amount on the left; 


next, set 1 year, $100 and the 1 mo.|12 mo. 
interest, on the right. Divide the $102.50) $100 
product of the numbers on the right $2.50 

by the product of those on the left; 102501800000 (29447 


the quotient is the rate. 


Notge.—It may be well to remark that the proportion given in connection 
with the two foregoing rules, is applicable to those problems in Interest where 
the principal, the interest and the time are given to find the rate. 


AVERAGING ACCOUNTS. 


The rule which determines the just time to pay, in one 
payment, several debts due at different times, is called 
Equation of Payments, or Average. 


Exam. 1. On January 1, 1888, Jno. Dwyer bought of 
Wm. Prior 


A bill of drygoods amounting to $300 @ 2 months, 
A bill of hats amounting to 400 @ 3 months, 
And sundries amounting to 500 @ 4 months, 


at what time may the whole be paid without loss to either 
party ? 


Nore. — The time which elapses before a payment is due is called the 
Term of Credit; and each item of a book account draws interest from the 
time it is due, which may be either at the date of purchase, or after a speci- 
fied time of credit. 


To find the average date of the foregoing account, we will as- 
sume that each item was due at the date of purchase; in other 
words, that they were cash transactions. 

On this assumption, it is evident that the purchaser would owe 
the merchant, at the end of each term of credit, not only the 


138 AVERAGING ACCOUNTS. 


amount of each bill of goods, but also the interest on each amount 
for the time, thus: 


Amount of account $300, int. for 2 mos. @ 6% = $3.00 
66 400, 66 3 66 66 pa 6.00 

as es 500, (4) a = 10.00 
$1200 ; $19.00 


The transactions were not for cash, however, but on time. The 
purchaser, therefore, is entitled to the $19 interest; in other 
words, he is entitled to hold the $1200 for such time after January 
1, us it would take that sum to give $19 interest at 6% per annum. 
The question, now, is, in what time will $1200 principal give $19 
interest, if $100 principal give $6 in 12 months? The propor- 
tion, or analogy, for this problem, would be: 


$1200 : $100 :: 12 mos. 
OD 
and this, by cancellation, becomes: 
es gee BS eee | 


Then, ee the second and third terms, and dividing by 
the first, we have 19 + 6 = 34 months, or 3 months and 5 days. 

If, then, the purchaser paid the entire $1200 of account, 3 
months and 5 days after January 1, there would be no injustice 
on either side. . 

Counting 3 months and 5 days from January 1, we find the 
average date to be April 6th. 


Exam. 2. On January 1, 1889, a man gave 3 notes, the 
first for $500 payable in 80 days; the second for $400 
payable in 60 days; the third for $600 payable in 90 
days. What is the average term of credit, and what the 
equated time of payment ? 


AVERAGING ACCOUNTS. 139 


Process. 
Interest on $500 for 30 days @ 6%-— $2 50 
“ AHO bao Ce es, 224-200 
% GOO) 4*5.- 90 Be = 9 00 
$1500 | G15 50 


Here, the interest of the several payments for the respective 
terms of credit is $15.50, and the sum of the payments, $1500. 
Now, in what time will $1500 give $15.50 interest, if $100 give 
$6 in 12 months? The following is the analogy: 


$1500 : $100 :: t& mos. 
Moe Li DO: 2, 


And, by cancellation, 6 disappears and 12 becomes 2. Then 
we have $15.50 multiplied by 2, or $31, which is multiplied in 
turn by 100, giving $3100; dividing this last, now, by $1500, 
gives 2; months, or 62 days, the average term of credit; and the 
equated time of payment, March 4. Hence the following 

Ruue. Find the interest on each payment for its term of credit at 
6 per cent., and add the results ; multiply double the interest thus 
found by 100, and divide the product by the sum of the payments; 
the quotient will be the average term of credit in months. 


Notse.— If the sum of the payments be not contained in the product as 
_ found above, multiply said product by 80, then divide, and the quotient will 

be the average term in days. 

The foregoing is called a Simple Equation, or Simple 
Average, having reference only to one side of the ac- 
count; and the terms of credit begin at the same date. 

When both debits and credits, or both sides of an 
account, are to be considered, and the terms of credit 
begin at different dates, the process is called a Compound 
Equation, or Compound Average; as illustrated in the 
following Ledger account: 3 


‘ 8 | v CO | ScF 


89 | IT |; 00 | 008 
89 | IL |} 00 | 008 ACR a hie eed 
00 | 0 00 | S&T 0 3 96 ‘AON 
iho 00 | 08% Pes 2 ” CL 390 
10 | F 00 | 08% ns “s {. 4ded 
Gg 13 00 | 008 sre Le og ydeg |! 0g | 9 00 | 0gE mo SOE 7 Oe o> 
f eg | 8 00 | 00¢ skep 901 | 93°AON 0} | ZI “SN || 09 | F 00 | OF skBp CIL | 93 °AON 0} | G  “SNY 
& 
as) . 
3 “ysoloqyUT]| “SJUNOUTY ‘OUILY, “qselojzuy!) “syunowWYy “OUL I, 
cal 
> 
<q ‘SsHOOUg 
QA 
a JOJVP OSBIOAV OY} ST JVYM SoAOqv SB JUNODOY 8,90Y PABLO puy aM “IaspaTy oy} sutusdg 
jo) 
jae 
e| 
S 
Ai Rete [eos tals a Shonen Sm ee LS alcegis 
00 | 08g ain weer cere reese ae Ie a ase 
00 08% wieake a: (alihiel tosh ele) & isliesipiteL eve. 616 re a I ‘ydag 
GOND00R ar eh ae amo rl a, ae OR a EOS COD. OCR Ts emia: Nan Cami ates gautiagy a OE - 
00 00g AS EN) ein cae aaa kb seg BO? | ZL ‘Sny 00 OFZ eeeececece ote She ae Oa. @ ‘Suy 
8gsl *Sasl 
‘ao ‘HOU GUVHOIY ‘ad 


140 


’ AVERAGING ACCOUNTS. 141 


First, assuming that each item was due at the date of purchase, 
and that Mr. Roe wished to settle his account on November 26 
(the latest date of the account), it is evident that each item bears 
interest from its respective date to the date of settlement, and that 
on November 26 he owed $1258 of account and $16.52 interest, as 
shown in the table opposite. Next, we find that he is credited 
with $500 and $300; he is now entitled to interest on each item 
from its respective date to November 26, the date of settlement, 
as shown in the table, viz.: $11.68. He owed the difference be- 
tween .both sides of the account; that is, $458 of account and 
$4.84 interest. He has, therefore, had the advantage of such 
time as it would take $458 to give $4.84 interest. The question 
now is, when should the account have been settled so that no in- 
terest would accrue? At such time before November 26 as it 
would take $458 to accumulate $4.84 interest at date of settle- 
ment. Multiplying double the interest, $4.84, or $9.68, now, by 
100, we get $968, which is divided by $458, the balance of ac- 
count, to get 2 months and 3 days, or 63 days. Counting 63 days 
back from November 26, we find the average date to be Septem- 
ber 28. 


Notrs.—1. If the balance of interest were on the credit side of the 
account, we should count forward from November 26th, as it is evident Mr, 
Roe would, in that case, be entitled to the time. 

2. Had there been a term of credit on the above account, say 2 months, 
for instance, then, the required date would be September 23d, plus 2 
months, or November 23d, 

3. If the terms of credit differed; that is, suppose each item in the above 
account was on 80 days’ time, except that of September 1, on which there is 
aterm of, say, 83months. In that case we would not take November 26th 
as the date of settlement, but the date on which the item having the longest 
term of credit falls due, which would now be December 1. On the other 
hand, allowing the dates to be as in the account, except that on September 
80th, Mr. Roe gives his note for, say, 2 months, instead of cash; then the 
maturity of said note, that is, December 3d, would be the date to which the 
interest on the account would be computed. The date to which interest is 
computed, in such cases, is called the focal date. 


142 AVERAGING ACCOUNTS. 


4, There may be such a combination of debits and credits, that the aver- 
age date will be earlier or later than any date of the account. 

5. The rate is optional, in computing interest on the account, but the same 
rate must be used for both sides of the account. The 6 per ceut rule, and 
60 days basis, is, perhaps, preferable to any other. 


Hence the 

Rue. (1.) Pind the interest of each item on the debit and credit 
sides of the account, at 6 per cent., from each respective date to the 
latest date on the account, whether such date be on the debit or credit 
side; the difference will be the balance of interest on the account: 
(2.) multiply double the balance of interest by 100, and divide the 
product by the balance of account; the quotient will be the time to be 
counted Backward from the focal date if the balance of interest and 
the balance of account be on the same side; but Forward if the balances 
be on opposite sides. (See note 8 to foregoing analysis.) 


Notrs.— 1. It is scarcely necessary to remark, that, in averaging an ac- 


count, we use the business method in computing the interest, namely: reject _ 


the cents in each item when less than 50, and when 50 cents, or more, add a 
dollar. The balance found from the items thus used will be the divisor in 
such cases. . 

2. In getting the interest for days, we take 60 days’ interest as the basis; 
thus, to find the interest of $240 for 115 days (the first item on the Dr. side 
of the table), $2.40 is 60 days’ interest; then 2 times $2.40, or $4.80 is 120 
days’ interest. From this we subtract 5 days’ interest or als of $2.40, which 
is 20 cents, giving $4.60, the interest for 115 days, and so with the remain- 
ing items. 

8. If the number of days be large, divide by 60; multiply the interest for 
60 days by the quotient, and add, for the remainder, the proper proportions 
of 60 days’ interest. Thus, what is the interest of $240 for 243 days at 6 


per cent? 
94/3 + 6]0 


Lt 
Dividing 243 days by 60, gives 4 for quotient, and 3 for remainder. Mul- 
tiplying $2.40 (60 eye interest) by 4, gives $9.60, or 240 days’ interest; 
then, 3 days equal >! y of 60 days, or 12 cents, making $9.72, the interest oe 
243 days. Again, the interest of $240 for 316 days, for instance, would be 5 
times 60, plus 16 hgh (31/6 + 6/0 =5..16). For 16 days we would say 10 = t 
of 60; and 6 = ao: the sum of the results is 16 days’ interest, etc. 


Bivorie ANID COss. 


Profit and Loss are commercial terms, having reference 
to the gain made, or the loss sustained, in the course of 
business. 

Gains and losses are usually estimated at some rate pet 
cent. on the money first expended or invested. 


Note.—It should be particularly remarked, that, by the gain or loss per 
cent, is to be understood the sum that would be gained or lost at the given prices, 
not on a hundred dollars’ worth sold, but on a@ hundred dollars laid out tu 
Jjirst cost, and in charges, if there be any. 


Ruel. The first cost and the selling price being given to find the 
gain or loss per cent.: As the first cost is to the gain or loss on 
that cost, so is $100 to the gain or loss per cent. 


Exam. 1. If tea be bought at 40 cents and sold at 50 
cents per pound, what is the gain per cent. 


Ais 1) To) 
4/0) L00|0 
25% 
Here, 50 — 40 — 10 cents, the gain on 40 cents; then, as 40 


cents (the first cost) : 10 cents (the gain on 40 cents) :: $100 (re- 
garded as first cost) : 25, the gain on $100. 


144 PROFIT AND Loss. 


Rue II. To find how a commodity must be sold to gain or lose a 


certain rate per cent.: As $100 is to the gain or loss on $100, or 


per cent., so is the first cost to the gain or loss on that cost; and 


from this and the first cost, the selling price will be found by 


addition or subtraction. 


Exam. 2. How must tea which cost 60 cents per pound 
be sold to gain 20%? 


LOO:2° 20.52.60 


——— eee 


12/00; then, 60 + 12 — 72 cents. 


Here, $100 (regarded as first cost): $20 (the gain on $100) 
.. 60 cents (the first cost) : 12 cents, the gain on 60 cents; then 
60 plus 12 — 72 cents, the required selling price. (By cancellation, 
We have 60 + 5 = 12.) Or, say: As $100 is to $100 plus the gain, 
or minus the loss per cent., so is the cost to the selling price. 'Thus, 
taking the same example, we have: 


1003 "1296.45260 
72|00 


Here, what cost $100 is sold for $120; and what cost 60 cents © 


will be sold, in proportion, for 72 cents. 


Rue III. Zo jind the first cost from the gain per cent., and the 


seluing price: As $100 plus the gain, or minus the loss per cent. 
is to $100, so is the selling price to the first cost. 


Exam. 8. Sold 12 musical instruments for $1500, and 
gained 25%, what was the first cost of each ¢ 


125 :. 100 :: 1500 
8 150000 
1000  1200|000 + 12 — $100 


PROFIT AND Loss. 145 


Here, the analogy is simply this: As the selling price, $125, 
is to $100 (considered cost), so is $1500, the selling price, to the 
cost price ($1200). ‘ 

To divide by 125, the dividend and divisor are multiplied by 8, 
and the new divisor is 1000, by which we divide, getting $1200, 
the cost price of 12 instruments. Dividing this by 12 gives $100, 
the cost price of each. 


QUESTIONS WITH THEIR SOLUTIONS. 


1. In closing the Ledger at the end of a year, the Dr. 
side of the Mdse. account is $38750, and the Cr. side 
$46500 ; what is the gain per cent. ? 

The Dr. side of Mdse. represents the purchases, or cost 
price, and the Cr. side, the sales, or selling price; the 
difference of the two will be the gain or loss on the ac- 


count, 
Sales. Purchase. 


46500 — 38750 — 7750 gain. 


SOP Tous Se 00 
) TT5000 (20% 
775000 


Here, we find the gain on the account to be $7750; then as the 
cost price is to the gain on that price, so is $100 to the gain per 
cent. (202). 


9. A merchant sold 24 musical instruments for $125 
each; he gained 25% on half, and lost 25% on the re- 
mainder; did he gain or lose on the transaction, and how 


much ? 
10 


a 


ANALYsIS: 12 at $125 each — $1500; then $100 plus the gain, 
$25, or $125, the selling price, is to $100 (considered cost), as 
$1500, the selling price, to its corresponding cost; thus: 


146 PROFIT AND Loss. 


195-100-322 1500 
8 150000 
1000) 1200/000 


Then, $1500, the selling price, minus $1200 cost = $300 gain | 
on the first half. 


Now, 100 — 98, or, 75 : 100 :: 1500 


4 150000 
3/00) 600000 
$2000 cost price. 
Then, $2000 cost minus $1500 selling price = $500 loss on 


the remaining half. 


The loss on the transaction is.... $500 
The gain f 6. eS eee ae 


Therefore he lost ...... | ore 2 8206 


3. A music dealer paid $1500 for 12 musical instru 
ments which he wished to sell at a profit of 20%; what 
must he charge for each instrument? 

i 
100 : 120 :: 1500 
pga 


This is a simple question in Compound Proportion, in which — 
$100, considered as cost, is to $120, the selling price, as $1500 


PROFIT AND LOSS. 14% 


cost is to its corresponding selling price; and next, as 12 articles 
is to 1 so is the price of 12 to the price of 1; the said proportion, 
by cancellation, becoming: 


10s 12271500 


and the fourth proportional $150, or the price which must be 
charged for each instrument. 


Nore. — From this we see that when the number of articles is 12, and the 
gain per cent. desired to be made is 20, we get the price of a single article, 
including 20 per cent., by simply dividing the price of 12 by 10. And to di- 
vide a number by 10 wesimply cut off one figure from the right of the divi- 
dend; in other words, move the decimal point one place to the left. Hence, 


When goods are bought or sold by the dozen we can readily 
tell what each article must be sold for so as to make 20 per cent. 
profit on the sale; and from this, by the method of aliquot parts 
(see page 109), we can get the price at any rate Pia cent., as illus- 
trated in the following: 


4. If a dozen silk hats be bought for $54, what must 
each be sold for to make 402 profit ? 


100 3140" 2354 
~ 45/60 + 12 
$6|30 


The process, by the regular method, would be: As $100, con- 
sidered cost, is to $140, the selling price, so is $54 cost, to its 
corresponding selling price, $75.60. Dividing this by 12, gives 
$6.30, the price at which each hat must be sold to make 402. 
Or thus: 

Dividing $54 by 10, gives $5. 40, the price of one hat, including 
20% profit, or the same as if the selling price were $120 ($100 
being considered cost). This lacks 20¢ of the desired profit. 


148 PROFIT AND Loss. ) 


20% is 4 of 120; now, by simply adding 4 of $5.40 to itself, we 
get the price at 40%, thus: : 


$5.40 — 120 
90 = 20 
$6.30 — 140 


Nortz.— A little practice will enable a person to solve such questions men- 
tally by this process. 

5. Suppose every thing as in the last question, only 10% 
was made by the sale, instead of 40; what was the sell- 
ing price of each hat? 


$5.40 — 120 
45 — 10 
$4.95 — 110 


10% is 5 of 120; subtract, the selling price is $4.95. 


Questions of the following nature will be found useful, 
and, from the examples and illustrations already given, 
will be readily understood : 


6. The population of Albany was 69422 in the year 
1870, and 90905 in 1880; what was the rate per cent. of 
the increase during the interval ? 

ANALYSIS: By taking the difference we find 21483, the increase 


of population. Then, 
As 69422 ; 214838 :: 100 : 30.94+, required rate. 


7. Between 1850 and 1870 the population of Albany 
increased by 36.76 per cent., and in the latter year it was 
69422; what was it in 1850 ? 

136.76 : 100 :: 69422 : 50762, nearly, the population 
in 1850. | 


DIVISION INTO PROPORTIONAL PARTS. 


Rute. To divide a given quantity into parts which shall be propor- 
tional to given numbers: As the sum of the given numbers is to 
any one of them, so is the entire quantity to be divided to the 
part corresponding to the number used as the second term of the 
proportion. 


Exam. 1. Proof spirits are composed of 48 parts of 
alcohol, or pure spirit, and 52 of water; how much of 
each is contained in 40 gallons of proof spirits? 


(48 + 52 — 100); then, 100 : 48 :: 40 : 19.20 alcohol. 
100 : 52 :: 40 : 20.80 water. 


Or, having found 19.20 alcohol, deduct it from 40; the differ- 
ence is 20.80 watcr, as found by the analogy. 


Exam. 2. Suppose a train to start from Albany to New 
York, going at the rate of 20 miles an hour, and another 
at the same time from New York to Albany, going 30 
miles an hour; where will they meet, the distance be- 
tween the two places being 145 miles ? 


150 Division INTO PROPORTIONAL PARTS. 


(20 + 30 — 50); then 50: 20 :: 145 : 58 miles from 
Albany. Or, by taking : 
the train from NewYork: 50: 80 :: 145; 87 miles from 
New York. 


The operation is proved to be correct by adding the results 
together. 


Exam. 8. Divide $9500 among father, mother and son 
in such a manner that the father’s share may be one-half 
greater than the mother’s, and the mother’s one-half 
greater than the son’s. 


ANALYysIs: Here the parts are evidently 1, 14 and 24, or 4, $ 
and 9. Then . 


(4+6+9—19) 19:4 :: 9500 : 2000, son’s share. 
19 : 6 :: 9500 : 8000, mother’s share. 
19 : 9 :: 9500 : 4500, father’s share. 


Exam. 4. A quantity of flax seed being converted into 
oil, the result was found to be 329 pounds of oil and 649 
pounds of cake; how much oil is that to the bushel, and 
what per cent.; a bushel of seed being 56 pounds, and a 
gallon of oil 7% pounds? 


(329 + 640 — 969); then 969 : 329 :: 56: 19.013, 

the number of pounds of oil to 56 pounds, or, to the bushel. 
Now, 19.0138 + 74 __ 2.535, or 24 gallons nearly. 
Next, 969 : 329 :: 100 : 33.952, or 34% nearly. 


Exam. 5. Pure water is composed of oxygen and 


DIVISION INTO PROPORTIONAL PARTS. 151 


hydrogen, in such proportions that the weight of the 
former is to that of the latter as 15 to 2. Required the 
weight of each contained in a cubic foot, or 1000 ounces, 
avoirdupois weight of water. 


(15 + 2—17); then 17: 15 :: 1000: Bey oz. oxygen. 
17: 2:: 1000: 11744 oz. hydrogen. 
7000 01 aie: 


15000 + aly 
900/00 + 102 
18 00_ 


882), TOT Si = i 


To divide the product of the second and third terms of the 
first proportion, or 15000, by 17, we multiply both by 6 (seeing 
that 17 is nearly one-siath of 100). Then dividing 90000 by 102 


ae II, Division), we get 882,38 ounces oxygen, the fraction 
qoz being equal to =5;. Now, 1000 — 88275 = 11744 ounces 
hydrogen. Or thus: 


120|00 + 102 
2/40 
117/60 
6 
66 + 6 == 11, that is, 11, 


Multiplying 17 and 2000 (2 x 1000), each by 6, we have 12000 
to be divided by 102. The quotient is 117, and the remainder 
66, which is divided py 6 (by which we Ma aa and the true 
remainder is 11, that is, when fully expressed, 44; or, ,6 = i. 


a. 
oe 
_—- a 


PARTNERSHIP. 


CasE I. 


The gains and losses of partners in business may be 
ascertained as in the last rule by the following proportion : 
As the whole stock is to the whole gain or loss, so ts the 
stock of any partner to his gain or loss. 


Exam. 1. A.and B. forma copartnership; A. furnishes 
$5000 and B. $7000 as capital; they gain $960; what is 
each man’s share of the gain ? 


(5000 + 7000 — 12000); then, 
12|000 : 5000 :: 960 
~ 4800/000 
$400, A.’s gain. 


B.’s gain is found by the same analogy, using 7000 for the sec- 
ond term. 


Exam. 2. Two brothers, John and James, purchase a 
house jointly for $25000; John contributed $10000 and 
James $15000 of the purchase-money. They let the 
house for the yearly rent of $2000; what share of the 
rent is each to receive ? 


PARTNERSHIP. 153 


25000 : 2000 :: 10000: 800, John’s share. 
25000 : 2000 :: 15000 : 1200, James’ share. 
$2000 


Case II. 


To find each partner’s share of the gain or loss when 
their capital is employed for wnequal periods of time. 


Rue. Multiply each stock by the time of its continuance in trade; 
then, using the products as stocks, proceed according to Case I. 


Exam. 1. A. and B. form a partnership. A. contrib- 
utes $3500 for 12 months, and B. $4500 for 9 months. 
They gain $1600; what is the share of each? 


3500 x 12 — 42000 

4500 x 9 — 40500 

82500 
Then, as $82500 : $1600 :: $42000 : $814.54, A.’s gain. 
82500: 1600:: 40500: 785.46, B.’s gain. 


The reason of the process will be evident from the consideration, 
that $3500 for 12 months is equivalent to 12 times that for 1 
month, that is, to $42000; and $4500 for 9 months is equivalent 
to 9 times that for 1 month, that is, to $40500. Hence, if these 
increased stocks be employed, it is evident that, since the times 
are then to be regarded as equal, the process will be the same as 
in Case I. 


Norte.— It need scarcely be remarked that the times, in all such operations, 
must be of the same denomination. If, for instance, one was 12 weeks, and 
the other 9 months in the foregoing example, the 9 months should be re- 
duced to weeks, or the 12 weeks to months. 


BANKRUPTCx 


The estate of a bankrupt may be divided among his 
creditors by the following analogy: As the sum of all 
the claims on the estate ts to the value of the whole estate, 
sv is the claem of any creditor to his dividend or share. 


Exam. 1. A bankrupt owes A. $350, B. $650 and C. 
$1500. His whole estate is worth only $1500; want is 
the share of each creditor ? 


($350 + $650 + $1500 — $2500) 


Then, as $2500 : 1500-:: 350 :°210, A.’s share. 
9500 : 1500:: 650: 390, B.’s share. 
9500 : 1500 : - 1500 « 900, C.’s share. 

$1500 — the whole ontater 


Nots. —In the division of a bankrupt’s estate, it is usual first to find how 
much on the dollar he can pay ; that is, how much the creditors will receive 
for each dollar of their respective claims. Thus, resuming the same ex- 
ample, we have this analogy or proportion: 

25/00 : 1500 :: $1, or 100 cents. 
1j00° 1500|00 
60400 

The sum of all the claims is to the whole estate as $1, or 100 cents, to the 
proportional part, corresponding to a dollar, which is found to be 60 cents, 
or 60 per cent, Then 60% of $350 — $210, A.’s share, as before. And B.’s 
and C.’s can be found in like manner. 


USEFUL RULES. 


Rue I. To find the rate at which a given principal will gain a 
certain interest in a stated time: 


As the given principal : $100 


Prk teed tine meet ae :: the interest : the rate. 


_ Exam. If $5000, invested for 1 year and 6 months, 
gain $525; what is the rate per cent.? 


The problem fully expressed is this: If $5000, in 18 mo. gain 
$525; what will $100 gain in 12 mo.; and for this we have the 
following proportion : 

As $5000 : $100 : : $525 : the rate. 


18 mo. : 12 mo. 


Reducing this to a simple proportion, we have the following : 
90000 : 1200 :: $525 : x, or the rate. Multiplying the second and 
third terms, now, and dividing by the first, we get the rate, thus: 

ca = 630000 + 90000 = 7%, the rate. 

Hence, if we draw a vertical line and set the 
given principal and the stated time on the left; 
and on the right, set the given interest, $100 and 5000/525 
1 year (or 12 mo. as the case may be); then divide ee ee 
the product of the numbers on the right by the 90000|630000(7. 
product of those on the left; the quotient is 
the rate. 


Norge. —It may be well to remark that, in all cases where the terms will 
admit, the work can be cut down by cancellation, 


156 Usrerut Rutes. 


Roz Il. To find what principal, in a stated time, will gain a cer- 
tain interest, at a given rate per cent, : 


As the given rate : $100 
the stated time : 1 year 


t : : the interest : the principal, 
Exam. What sum of money must be invested to gain 
$525 in 1 year and 6 months, at 7 per cent. ? 
The problem fully expressed is this: If $100 gain $7 (7%) in 12 


months; what sum will gain $525 in 18 months; and for this we 
have the following proportion : 


het : de : : $525 : the principal. 


And by reduction to a simple proportion, we have the following: 
As 126 : 1200 : : 525 : x, or the principal. 
Multiplying the second and third terms, and dividing by the first, 
we have: 10 <5 _ 630000 + 126 = $5000, the principal. 


126 
Or, drawing a vertical line, setting the given rate 155 
and the stated time on the left; and the given in- 18/100 
terest, $100 and 12 mo. on the right, and dividing the os 12 
product of these on the right by the product of those 126|630000 
on the left, we get the principal. $0000 


Rue Ill. To find the time in which, at a given rate per cent. per 
annum, a given principal will produce a certain interest : 


As the principal : $100 
the rate : the interest 


:: L year : to the time. 
Exam. How long will it take to have $5000 gain 
$525, at 7 per cent. per annum, simple interest ? 
The problem fully expressed is this: If $100, in 1 year, gain 


$7 (7%) how long will it take $5000 to gain $525; and for this we 
have the following proportion : 


Usrerut Rv ues. 157 


As ie ie :: l year : the time required. 


Reducing this to a simple proportion, we have the following: 
As 35000 : 52500 :: 1 : x, or the time. 


And we have, now, simply, 52509 to be divided by 35000. Cut- 
ting off the three ciphers in 35000 and three places from 52500 
(this divides each by 1000, and does not alter the proportion) we 
have 52.5 to be divided by 35, as shown in the margin. 


To divide by 35, we use the component factors, 5|52.5 
5 and 7 (5 X 7 = 85) the quotient is 1.5 years, or 710.9 | 
tcyT-,. 0-0, 1.5 years 


Now, the interest on $5000 for 1 year @ 74, is $350; and the given 
interest is $525, and if 525 be divided by 350 the quotient is the time; 
hence, the rule may be stated as follows: 


Divide the given interest by the interest on the principal for 1 year, 
at the given rate, and the quotient is the time in years and decimals. 


Rue IV. To find what principal, in a stated time, will increase to 
@ given amount,.at a given rate per cent. per annum: 


This is the same as finding the true present worth of a debt, and 
the rule given on page 133, will answer here, also, viz.: 


As $100 plus its interest for the given time, and at the given rate, 
is to $100, so is the given amouut to the principal sought; illustrated 
in the following : 


Exam. What principal, in 1 year and 6 months, at 
7 per cent., simple interest, will amount to $5525? Or, 
in other words, what is the true present worth of a debt 
of $5525, due in 1 year aad 6 months, at 7 per cent.? 

Here, the interest of $100 for.1 year and 6 months, at 7%, is $10.50. 
Then, $100 + $10.50 = $110.50, the amount of $100 for the given 


time and at the given rate; and applying the foregoing rule, we 
have the following proportion : 


£58 Usrrut Ru es. 


As $110.50 : $100 :: $5525 : x, or the required principal. Throwing 
off the decimal, now, in the first term, we have the following: 
11050 : 10000 :: 5525 = “WY <*> — 55250000 + 11050 = $5000, 
the required principal, or, the true present worth of the debt. 


Questions like the following, and which are of a useful kind, are 
of the same nature as those regarding the interest of money :‘ 


Exam. If the population of a city was 240000 in the 
year 1896, and 300000, in 1906; what was the rate per 
cent. of increase during the interval ? 

By taking the difference of these we find 60000, the increase of 


population. The question now is: If 240000 gain 60000; what will 
100 gain; and for this we have the following proportion : 


As 240000 : 60000 :: 100 : x, or the rate; that is: Oa = 
6000000 + 240000 = 25 per cent., the rate. 


Exam. Between 1896 and 1906, the population in- 
creased 25 per cent., and in the latter -year it was 
300000; what was it in 1896? 

In this, 100 + 25= 125: Then, we have 125 : 100 :: 800000 : x; 
100 x 300000 
——— = 30000000 + 125 = 240000, the popu- 


125 
lation in 1896. (See examples, page 148.) 


and this gives us: 


Exam. If $25000 be invested in property which rents 
for $3500 a year, and on which $500 are paid in taxes; 
what rate of interest does the investment pay ? 


Here, $3500 — $500 = $3000, the income from $25000. Now, 


$25000 : $3000 :: $100 : rate; and we have “~~ <" — 300000 + 25000, 


or, cutting off three ciphers from each, we have 300 + 25 = 122, 
the rate. 


PRACTICAL HINTS FOR BUILDERS, 


Lumber and sawed tember, as plank, scantling, etc., 
are usually estimated in Loard Measure, hewn and round 
timber in cubic measure. 

A board foot is 12 inches long, 12 inches wide and 1 
inch thick ; in other words, it isa square foot 1 inch thick. 

In board measure all boards are assumed to be 1 inch 
thick. 


Notre. — Lumber 1 inch thick or less is sold by surface measure, and in 
the trade is denominated boards. Ifmore than 1 inch thick it is called plank, 
and is computed at 1 inch thickness, or standard thickness; that is, the 
product of the surface measure in square feet multiplied by the thickness in 
inches is the number of feet of lumber at board measure. 


Rue. To jind the number of feet, board measure, in a board or 
plank: Multiply the length in feet by the width in inches and 
divide the product by 12; the result is the number of feet at 1 
inch, or the standard thickness. Next, multiply the result thus 
found by the thickness of the plank in inches; the product is the 
number of feet of standard thickness, or board measure. 


Exam. 1. How many feet board measure in a piece of 
pine lumber 16 feet !oug, 9 inches wide and 1% inches 
thick ? | 


160 PRAcTICAL HINTS FOR BUILDERS. 


ANALYsiIs: 16 X 9 = 144; then, 144 + 12 = 12 feet at 1 
inch thick. 

Then, 12 x 14 = 15 ft.; or simply add 4 of 12, or 3 
and we have the number of board feet = ‘15 feet. 

If the plank were 14 inches thick we would add the $ of 12, or 
6, making 18 feet, board measure; if 24 inches thick, it would be 
24 times 12, or 27 feet, etc. 


Note. — When the piece is 12 feet long and 1 inch thick, or less, the sur- 
face feet will be the same as the width in inches; thus, a buard 12 feet long, 
7 inches wide and 1 inch thick is 7 feet, surfaee measure. 


In the trade it is customary to have 1} inch lumber resawed into 
boards about 4 inch thick, commonly called panel-stuff, which is 
bought and sold by surface measure as if it were inch, or stand- 
ard thickness; but the price is reduced accordingly. For in- 
stance, 1} inch pine lumber, worth $50 per thousand feet, would, 
when resawed, be sold for about $32 to $35 per thousand feet. 

Hence, questions of the following nature are frequently asked: 


Exam. 2. Whether isit more advantageous to buy 1000 
feet of 14 inch pine at $50, and have it resawed into 
panel-stuff, paying $2 for the sawing, or to buy the same 
quantity already resawed, at $35 per thousand ¢ 


1000 feet of 14 inch @ $50, and $2 for sawing — $52. 


In 1000 feet of 14 inch measurement there are 800 feet at 1 
inch, or surface measurement, found thus: 


1000 feet of 14 
Jess one-fifth (4) 200 
800 feet, 


There are 5 quarters in 14, therefore, 1 quarter (4) is the fifth 
of 5; deducting this leaves the 1 inch, or surface. Now, since 


Practical HINTS FoR BUILDERS. 161 


every piece of 1} inch makes 2 pieces when sawed, 800 feet is 
doubled, giving 1600 feet of panel. Then 1600 feet at $35 per 
thousand gives $56, or $4 more than $52, as found above. It 
would be more advantageous, therefore, to buy at $50 and pay $2 
for sawing. 


Exam. 3. A carpenter wishing to get 1000 feet of 1- 
inch pine boards dressed to ¢ inch, and not getting the 
quality suitable for his purpose in the market, concluded 
to take 14-inch lumber and have it resawed ; how much 
of the latter did he require ? 


To solve problems of this nature, it must be borne in mind that 
the surface is the same regardless of the thickness. 

But 1000 feet surface is 1500 feet, standard measure, at 14 inch 
thick, and since each piece which goes to make 1500 feet will 
make two pieces when resawed, it follows that half the quantity 
will make 1000 feet surface. 

Dividing 1500 feet then by 2 gives 750 feet of 14-inch lumber, 
the required quantity. 

Proor: 750 feet 14-inch lumber, 
less 4 250 (1 half inch = 4 of 3 half inches, or 14 inch) 

500 feet at 1 inch thick, and doubling this gives 1000 
feet surface; that is, 750 feet of 14 inch resawed. 

Or the problem might be solved thus: take half 1000 feet sur- 
face and we have 500; to this add one-half and we have 750 feet 
of 14, inch as before. Adding one-half of 500 feet to itself, we 
need scarcely remark, is multiplying 500 feet surface by 14, the 
thickness. 


SouTHERN PInp. 


Southern pine, commonly called ‘‘ Georgia pine,” or ‘‘ yellow 
pine,” comes in various lengths and widths, and it not unfre- 
quently happens that builders requiring a quantity of flooring or 


1) 


162 PrRacticAL HINTS FOR BUILDERS. 


ceiling will give a hurried order to the dealer for such quantity; — 


and the dealer, in delivering the same from the mill, where it has — 


probably just been dressed, will not spend the time to get the 
standard measurement, but, instead, will merely take the lengths 
of the pieces, or the lineal feet, converting the same at leisure 
into standard, or. board measure. For such emergencies we give 
the following simple 


Rue. To jind the lineal feet for any surface and to convert the 
same into standard or board measure: Multiply the surface to be 
covered by 12, and divide the product by the width of the board 
or plank; the quotient will be the lineal feet. Next, reverse the 
process; that is, multiply the lineal feet by the width of the board 
or plank (in the rough, or before being dressed), and divide by 12; 
the result is surface feet. Then add for the extra thickness, if 
more than 1 inch thick, and we have the standard, or board 
measure. 


Exam. 1. How many lineal feet of 14 by 4 inches (face) 


of Southern pine will cover 2550 feet surface 2 
9550 x 12 + 4 — 7650 lineal feet. 
Reason of the rule: When a board or plank is 12 feet long it 


will cover as many feet surface as there are inches in the width 


of the face of such board or plank. Now, if a piece 4 inches on 
the face and 12 feet long cover 4 feet surface, how many feet in 
length will cover 2550 feet surface? And for this we have the 
following proportion, or analogy: 4: 2550 :: 12; that is, 4 feet 
surface is to any given surface (2550 in this case) as 12 feet (the 
length corresponding to 4 feet surface) is to the length, or lineal 
feet, corresponding to 2550 feet surface. Multiplying the second 
and third terms and dividing by the first, we get 7650 feet in 
length, or lineal feet. 

Next. how many feet, standard, or board measure, in 7650 lineal 
feet of 14-inch flooring, 4 inches on the face ? 


PRACTICAL HINTS FOR BUILDERS. 163 


A piece which gives 4 inches (face) when dressed, is usually 43 
inches in the rough; that is, before being dressed, and, as it és 
customary to buy and sell at what the lumber measures in the rough, 
we multiply by 44 instead of 4, thus: 

' 7650 < 44 = 2868 feet, the surface, 
12 
then adding} (the thickness over linch)or 1717, we find the 
standard, or board measure to be 3585 feet. 
Hence, in 
MAKING ESTIMATES 


for flooring, ceiling, etc., the calculations for the material should 


be made on the measurements in the rough. 


Exam. 2. How many feet of “ Georgia pine ” flooring, 
24 inches on the face and 14 inches thick, required to 
cover 3 floors 40 x 386 feet ? 


40 x 86 x 3 = 4820 feet, the surface to be covered. 


Now, a piece of flooring 24 inches on the face and 12 feet long 
(regardless of the thickness) wil cover 24 Jeet surface. But a piece 
vf flooring 24 inches (face), 12 feet long, was 3 inches (face) in the 
rough, or 3 feet surface, and at 1} inches thick it was 33 feet, 
board measure. 

The question now is: if 3% feet, board measure, cover 24 feet 
surface, how many feet, board measure, will cover 4820 feet sur- 
face? And the proportion is: 


21 + 4390 :: 38 


Or, reducing the first and second terms to the same denomina- 
tion, halves, and the third to quarters, or fourths, 


~ we have 5 : 8640 :: 15; and by cancellation 
this becomes 1 : 8640:: 3. 


164 PRACTICAL HINTS FOR BUILDERS. 


Multiplying 8640 by 3, now, gives 25920 fourths, or quarter 
feet; which, being divided by 4, gives 6480 feet, board measrre, 
in the rough, or the number of feet to be paid for. From this, it 
will be seen that, when the material is narrow, it takes about 14 
times the surface to be covered, for the required number of feet, board 
measure, at 14 inches thick. 


Exam. 3. A builder requires 4 inch pine ceiling, 24 
inches on the face, to cover 10000 feet surface. He can 
buy ceiling suitable for his purpose for $30 per thousand 
feet, surface measure, or, 14 inch by 6 inch pine, from 
which to make such ceiling, for $35 per thousand feet. 
Which is the more profitable, the cost for making each 
piece of ceiling from the 14 inch being 23 cents? 


AnaLysis: A piece 24 inches on face, 12 feet long (regardless 
of thickness), will cover 24 feet surface. Dividing 10000 feet by 
24, or using their doubles, 20000 + 5, gives 4000 pieces at 24 feet 
each. Buta piece 24 feet surface, dressed, was 3 feet in the 
rough, and, therefore, 4000 pieces at 3 feet each, would make 
12000 feet surface measure, or what has to be paid for. This, at 
$30 per thousand, is $360, the cost. 

Next, a piece 14 by 6 inches and 12 feet long contains 6 feet 
and the quarter of 6, or 74 feet, standard measure. It will take 
1000 pieces of 14 < 6 (each piece makes 4 pieces of ceiling when 
milled) to make 4000 pieces of ceiling. Now, 1000 pieces at 74 
feet each makes 7500 feet, standard or board measure, and at $35 
per thousand this gives $262.50. Adding to this the milling of 
4000 pieces at 24 cents each, or $100, we have $862.50, the cost. 
It is more profitable to buy the ceiling already made, in this case. 


_ Norz. — From the three foregoing examples it will be seen that, by taking 
12 feet in length as a basis of calculation, estimates for flooring, ceiling, etc., 
ean be readily made. 


PRAcTICAL H1InNTs FOR BUILDERS. 165 


Roor ELrvaAtions, ETC. 


By the ‘‘ pitch ” of the roof is meant the ratio which the height 
of the ridge above the level of the roof-plates bears to the span, or 
the distance between the supports or studs on which the roof rests. 

The usual pitches are the Common or true pitch, in which the 
rafters are three-fourths of the width of the building; the Gothic 
pitch, in which the length of the principal rafters is equal to the 
width of the building; the Pediment pitch is when the perpen- 
dicular height is 2 of the width. There are also the 4 pitch, + 
pitch, % pitch, etc. 


Rue. To jind the length of rafter for any particular pitch of 
roof: To the square of the perpendicular height of roof add the 
square of half the width of the building, the square root of the 
sum is the length of the rafter. 


Nore.— The method of extracting the square root is given in almost any 
common-school arithmetic. 

It need scarcely be remarked that the rafters for the Common pitch, and 
also for the Gothic, are obtained without the aid of this rule. 


SHINGLES, Latu, Ere. 


A ‘‘shingle” is 4 inches wide and from 16 to 18 inches long. 
But shingles are seldom made of a uniform width; they vary from 
2 to 10 inches, more or less, and are put up in bundles, or bunches, 
containing 250 shingles each (not by count but on an average of 
4 inches to a shingle). Hence, there are 4 bunches to 1000 
shingles. 

Since a shingle is reckoned at 4 inches, it is evident that the 
number of shingles required to cover a roof will depend on how 
much of the shingle is ‘‘laid to the weather.” Thus, if 6 inches 
be laid to the weather, a shingle will cover 24 square inches 
(4 x 6); and by dividing 144 square inches (1 square foot) by 24, 
we find it will take 6 shingles to cover 1 square foot. 


ig Bed em 


166 PRACTICAL HINTS FOR BUILDERS. 


Again, if laid 5 inches to the weather, a shingle will cover 20 
square inches (4 x 5); and dividing 144 by 20, gives 74 shingles 
to the square foot. 

Hence, it will be seen that, by multiplying the number of 
square feet to be shingled, in the one case, by 6, and by 74 in 
the other, we get the number of shingles required at 6 inches and 
5 inches, respectively, to the weather. } 


Now, shingles are generally laid from 5 to 54 inches to the © 


weather, and for practical purposes the following simple rule will 
be found sufficiently accurate: Multiply the number of square feet 
to be shingled by 7; the product is the number of shingles required, 
nearly . 


Exam. How many sawed pine shingles required to 
cover a building 50 feet in length and 36 feet in width, 
the roof being of the common or true pitch ? 


ANALYSIS: In the true pitch the rafter is ¢ of the width of the 
building; % of 36 = 27 feet, the length of the rafter. Then, 27 
doubled and multiplied by 50 will give the surface, or the num- 
ber of square feet to be shingled, thus: 

27 X 2 X 50 = 2700 feet, the surface of roof; and 2700 X 7 = 
18,900 shingles, the number required. 


LatH. 


Notr. — It is customary among the dealers to make use of the singular 
form, lath; as, ‘‘Have you got any lath?”’ ‘‘How many lath will cover 
1000 feet?” etc. 


A lath is 4 feet long, 14 to 1$ inches wide and about % inch 
thick, usually made from pine, spruce or hemlock. 

Lath are seldom, if ever, counted in bunching; they are gen- 
erally put into a gauge, or measure, which contains about 100 
pieces, more or less, and tied in a bunch; hence, 10 bunches 
make 1000 lath. 


on ae 


PRACTICAL HINTS FOR BUILDERS. 167 


The surface of a lath 4 feet, or 48 inches long, and 14 inches 
wide, is 48 X 14, or 72 square inches, and of 2 lath, 72 xX 2, or 144 
square inches: therefore, 2 lath, 14 inches, set edge to edge, will 
cover a square foot. Hence the following simple 

Rute. To find the number of lath required to cover any surface: 
Multiply the number of square feet to be lathed by 2; the product 
is the number required. 


Exam. How many lath will be required for a room 24 
feet long, 20 feet wide, and 9 feet 6 inches high ? 


ANALYSIS: The length of the four walls is (24 + 24 + 20 + 20) 
88 feet; then 88 x 94 (the height 9 feet 6 inches) — 836 square 
feet, or surface of walls. Next, 24 x 20 (the ceiling) = 480 square 
feet or surface of ceiling. Putting the two surfaces ~—— 
EY GOES til al SO eee Poh Ten rar cena 1316, ..the 
number of square feet to be lathed. Doubling this, we have 2682, 
the number of lath required. 


Nores.—1. Lath are usually set about 1¢ inch apart, to allow for the 
“‘clinch’’?; hence, when 11g inch lath are used, a deduction of about one- 
tenth, or 1 bunch in every 10, may be made. Thus, in the foregoing 
example, 2632 less 263 (345 of 2632) — 2369, would be nearest the true re- 
sult. But for 144 inch Jath no deduction is necessary. 

2. Allowance must, of course, be made for doors, windows, ete. 


CoNSTRUCTION AND CAPACITY OF Bis, ETC. 

The Standard Bushel of the United States is a cylindrical meas- 
ure 184 inches in diameter and 8 inches deep, and contains 2150.42 
cubic inches. (The capacity, 2150.42, is found by multiplying 
the square of the diameter, 184, or 18.5, by .7854, and the pro- 
duct by the depth, 8 inches.) 

Since a cubic foot contains 1728 cubic inches, and a standard 
bushel contains 2150.42 cubic inches, a bushel is equal to 14 cubic 
feet, nearly (2150 + 1728 = 1}, nearly), the proportion being 1 
to 14, or 4 to 5, nearly. Hence, 


168 PRACTICAL HINTS FOR BUILDERS. 


To jind the capacity of a bin in bushels: Add 4 of 
the quantity in bushels to itself; the sum will represent 
the capacity of the bin. 


Thus, what must be the capacity of a bin to contain 160 bushels 
of wheat? 160 + 40 G4 of 160) = 200, the number of cubic feet, 
or capacity of bin. And 


To jind the number of bushels contained in a bin: 


Deduct 4+ of the capacity of the bin from itself; the re- 


mainder will represent the number of bushels. 


Thus, how many bushels of wheat in a bin of 200 cubic feet 
capacity? 


200 — 40 (4 of 200) = 160 bushels of wheat in a bin of 200 | 


cubic feet. Hence, 


Any two dimensions of a bin being gwen, the third 
dimension can be found, thus: 


(1) Increase the number of bushels by 4 of itself; the result will 
represent the number of cubic feet contained in the bin. (2) Di- 
vide the contents in cubic feet by the product of the two dimen- 
sions, and the quotient will be the other dimension. 


Exam. 1. What must be the depth of a bin to contain 
280 bushels, its length being 10 feet and its width 5 feet ? 


ANALYSIS: 280 + 70 = 350; then 350 + 50 (10 x 5) = 7 feet, 
the depth. 


Exam. 2. What is the value of a bin of wheat 20 feet 
long, 12 feet wide, and 5 feet deep, at $2 a bushel? 


PRACTICAL HINTS FOR BUILDERS. 169 


ANALYsIS: 20 x 12 x 5 = 1200; then, 1200 — 240 (4 of 1200) 
= 960, the number of bushels in bin. 960 at $2 = $1920, the 
value. ; 


Exam. 3. A coal bin is 12 feet long and 6 feet wide; 
how deep must it be to contain 12 tons of chestnut coal, 
the contents of a ton of chestnut coal being 38 cubic feet? 


ANALysIs: 38 x 12 = 456, the contents of 12 tons; then 456 + 
72 (12 x 6) = 64 feet, or 6 feet 4 inches, the depth. 

Rue. To find the capacity of a vessel or space in gallons: Divide 
the contents in cubic inches by 231 for liquid gallons, or by 268.8 
for dry gallons. 


Exam. 1. How many gallons of water will a cistern 
hold that is 4 feet by 5 feet, and 6 feet deep ? 


Anaxysis: (4 X 5 X 6 X 1728) + 231 — 89734 gallons capacity. 


Exam. 2. What must be the depth of a cistern that is 
6 feet long and 54 feet wide to hold 462 gallons of water ¢ 


52 x 231 
ANALYSIS: iaresis = 1.87 feet, the depth. 
Exam. 3. A cellar 40 feet long, 20 feet wide and 8 
feet deep is half full of water. What is the cost of 
pumping it out at 6 cents a hogshead ¢ 


Anaxysis: (40 X 20 X 8 X 1728) + 231 — 47875.32 gallons; 
then 47875.32 + 63 (gallons in a hogshead) = 759.92 hogsheads; 
and 6 times 759.92 — 4559.52 cents. Dividing this by 2 (half the 
cellar) gives $22.80, the cost. 


1,0 PRACTICAL HINTS FOR BUILDERS. 


Masonry. 


Masonry is estimated by the cubic foot and by the perch, also by 
the square foot and the square yard. 

A perch of masonry is 164 feet long, 14 feet ie and 1 foot 
high. Multiplying these three dimensions together, we find there 
are 242, or 24.75, cubic feet in a perch of masonry (164 x 14 <1 
= 24%), or (16.5 x 1.5 X 1 = 24.75). 


Notr. — When stone is built into a wall without mortar or filling an allow- 
ance of 234 feet is made, and 22 cubic feet make a perch. 


Rute. To find the number of perches of masonry ina wall: Divide 
the number of cubic feet in the work by 24%, or 24.75; the quo- 
tient will be the number of perches. 

Nots.— Brick-layers and masons, in estimating their work by cubic meas- 
ure, make no allowance for the corners of the walls of houses, cellars, etc., 
but estimate their work by the gzrt, that is, the entire length of the wall on 
the outside. 


Joiners, brick-layers, and masons, make an allowance of one-half the 
openings or vacant spaces for doors, windows, etc. 


Exam. 1. At $4 a perch what will be the cost of build- 
ing the walls of a cellar 374 feet long, 26 feet wide, 9 feet 
deep and 2 feet thick ? 


ANALYSIS: 75 + 52 = 127 feet, the girt, or outside measure; 
then, 127 x 9 x 2 = 2286 cubic feet, the solid content of walls. 
Next, 2286 + 24.75 = 92.36 perches. Multiplying this, now, by 
4, gives $369.44, the cost. 

Nore.— To divide by 24.75, or 2484, we make use of the simplified method, 


as pointed out at page 60, example 4, to which the reader is referred. 


EXCAVATING 
or digging is measured and paid for’by the cubic yard, and a cubic 


yard of earth is called a load. 


Nore. — In a lineal yard there are 3 feet. Cubing 3 (3 X 3 x 8 = 27), we 
get 27 cubic feet for acubic yard. Hence, a box 9 feet long, 8 feet wide 


PRACTICAL HINTS FOR BUILDERS. id 


and 1 foot deep will contain a load of earth (9 x 3 « 1 = 27). And any two 
dimensions of a box to contain a load being given, the other dimension can 
be found, thus; Divide 27 by the product of the two given dimensions; 
the quotient is the other. (See example 1, page 168.) 


Exam. 2. What is the cost of digging the cellar in the 
last example at 50 cents a load? 

ANALYSIS: 874 X 26 X 9 = 8775 cubic feet of earth in cellar. 
Then, 8775 ~ 27 = 825 loads; and at 50 cents a load it is $162.50, 
the cost. 

Brick-work. 

Rute. Zo find the number of bricks in a wall: Multiply the 
number of cubic feet in the work by the number of bricks ina 
cubic foot; the product is the number of bricks required. 


Nore. — About 22 common bricks make a cubic foot when laid. 


Exam. 1. How many common bricks in a wall 70 feet 
long, 20 feet high, and 12 inches thick ? 


Anatysis: 70 x 20 x 1 = 1400 cubic feet in wall; then 
1400 x 22 = 30800 bricks. 

In estimating brick-laying by the square yard, the rod, or by 
the square of 100 feet, the work is understood to be 12 inches, or 
14 bricks thick, which is called standard thickness, 

Rue. To reduce brick-laying to standard thickness: Multiply the 
superficial content of the work by the number of half bricks in 
thickness, and divide by 3. 

Notge.— The superficial content is found by multiplying the length by the 
height. <A rod is 161g feet long, consequently a square rod is 272.25 square 
feet (16.5 X 16.5 = 272.25). 


Exam. 2. How many squares of brick-work in a wall 
130 feet long, 12 feet high; the wall being 24 bricks 
thick ¢ 

ANALysis: 130 X 12 = 1560 feet, superficial content, and 
1560 x 5 = 7800; then, 7800 + 3 = 2600 feet of standard work, 
or 26 squares (2600 + 100 = 26). 


172 Practica, Hints ror BuILpErRs. 


LuMBER CALCULATIONS 


In computing the cost of lumber, or other merchandise, at so 
much per thousand, the desired results are, in most cases, obtained 
more easily by the method of aliquot parts, than by the usual 
methods of multiplication ; thus: 

Take, say, 15886 feet of lumber. 

At $1000 per thousand feet, the cost is at $15836 = $1000 
sight, or as many dollars as there are feet; $1583.6= $100 
and at $100, the cost is a tenth of that at $158.36—= $10 
$1000. At $10, the cost is a tenth of $100; $15.836—= $1 
at $1, a tenth of $10; at 10 cts., a tenth of $1.5836= 10 cts. 
$1 and at lct. itisatenth of 10 cts. Hence, .15886= a3 


In computing the cost of lumber, if we assume $100 per thousand, 
as a standard price, calculations can be simplified ; thus: 


Exam. What is the cost of 15836 feet of pine lumber 
@ $27.75 per thousand feet ? 


Cutting off one figure gives the cost at $100. 
Then, the cost at $25 is a fourth of that, or $1583/6 =$100 


$395.90; now, at $2.50, the cost isa tenthof — a era $20 
that at $25, or $39.59 and at 25 cts., a tenth 2 ks ay 


of that at $2.50,.or $3.959. Adding the “$139|449— $9775 
= 15 
several results gives the cost at $27.75. > ded 


Exam. What is the cost of 75680 feet of black walnut 
at $75.75 per thousand ? 


In this, we have at sight the cost at $100; 
then, at $25, it is a fourth, or $1892, which $7568|0 = $100 
is deducted, leaving $5676, the cost at $75. _ 1802/0) ae 
Now, 75 cts. is a hundredth part of $75, or ye we $75 m5 
$56.76, which is added; this gives $5732.76, $5733\76— $75.75 
the cost at $75.75. 


PracticaLt Hints ror BuILpERs. 173 


Exam. What is the cost of 75684 feet of lumber at 


$32.58 per thousand feet ? 


Here, we cut off one figure to get the cost 
at $100 ; then, one-fourth of this, or $1892.10, 
is the cost at $25 ; now, $5 isa fifth of this, or 
one-half of the cost at $10, got from the top 
line ; next, $2.50 is half of $5; and for 8 cts. 
we take 8 times the cost at 1¢, that is, .75684, 
got from the top number: say 8 times .757, 
making use of three figures only, and allow- 
ing for those rejected; the result is $6.056. 

Or thus: 

Cutting off two figures gives the cost at $10, 
always; then, 3 times 10 are $30; now, $2.50 
is a fourth of $10; and finally, 8 times .757 
gives $6.056; the sum of the several results 
gives $2465.786, the cost at $382.58, the same 
as before. 


$7568/4 —$100 


1892it “= + 25 
878/42 = 5 
189|\21 = 2.50 

6|056= .08 


$2465|786= $32.58 


$756/84 —$10 

2270/52 = 30 

189/21 = 2 50 
6\056=- .08 


$2465|786 = $32.58 


Exam. What is the cost of 75684 feet, at $37.75 per 


thousand ? 


In this, the cost at $25 is one-fourth of that 
at $100, or $1892.10; then, $12.50 is half of 
$25; and 25cts. is the hundredth part of $25; 
the hundredth part of $1892.1 is $18.92; the 
sum of the several results is $2857.07, the re- 
quired cost. 

Or thus: 


Here, we have the cost at $10, at sight, and 
3 times this is the cost at $30; then, $5 is half 
of $10; now, $2.50 is half of $5, and 25cts. is a 
tenth of $2.50; the sum of the several results 


is the cost at $37.75, as before, 


$7568|4 =$100 


189211 — 25 
946|/05= 12.50 
18/92= 25 


$2857|07= $37.75 


$756 84=$10 

3970 50-2 30 
8378/142— 5 
189/21— 2.50 
18 92— OD 


$2857|07=$37.75 


CHRONOLOGICAL CALCULATIONS. 


To find the weekly day for any given date from the 
year 1600, New Style (N. 8S.) for any year or century 
thereafter, we give the following simple 


RuLE |. Subtract the centuries from the given year. 
Il. To the remainder add one.fcurth of the given year; also the 


number of days from January 1, up to and including the given date, 


and 1 for every fourth century. 

III. Divide the sum by 7; the remainder, counting Sunday 1, will 
be the weekly day. 

Notes. —1. For leap years add 1 to the centuries before subtracting always, 
counting 29 days in February; then proceed according to the rule. 

2. When 7 is contained in the sum without remainder, Saturday is the weekly 
day. 

3. In taking the fourth part of the given year, the remainder, if any, may be 
rejected; also, in taking one-fourth of the centuries. 


Exam. 1. On what day of the week did the 4th of 
July, 1775, happen ? 


SoLUTION.— Subtracting 17 centuries. from 1775 Jan. 31 da. 
the given year, we get 1758; to this we add 17 Yeb. 23 
one-fourth of 1775, rejecting the remainder, 1758 Mch.31 “ 
next 185, the number of days from January 1, 443 April30 “ 
up to and including July 4th, and 4, 1 for 185 May 31 ‘* 
every fourth century in 17 ee rejecting 4 June 30 ‘ 
the remainder. 7)2390 July 4 “ 

Dividing the sum, 2390, by 7 gives a re- 341—3 185 
mainder of 3; then counting Sunday 1, Mon- 
day 2, Tuesday 3; we find that July 4th hap- 
pened on the third day of the week, Tuesday. 


CHRONOLOGICAL CALCULATIONS. 175 


Exam. 2. On what day of the week will Washington’s — 
birthday (Feb. 22) happen in 1912? 


In this, 1912 being a leap year, we sub- 1912 less (20) 19 +1 
tract 20 from-the given year, the remainder ~ 7899 
is 1892. Adding to this 478, the fourth part 478 
of 1912; then 53 days, the number from Jan- 53 
uary 1 to February 22, and 4 (19 + 4) 1 for 4 
every fourth century in 19 centuries, and 9427 
dividing the sum by 7, we obtain a re- Dill 
mainder of 5; the fifth day of the week, 346...5, Thursday. 
Thursday. 


To find the Day of the Month on which a Particular Day oy the 
Week will happen. 

Exam. 2. The presidential election occurs on Tuesday 
after the first Monday in November; what day of the 
Month will it be in 1912? 


To solve a problem of this kind, we have to find, first, on what 
day of the week November 1 will happen. 


SOLUTION.— 1912 being a leap year we sub- 1912 less 20 
tract 20 (19+ 1) from the given year, as in the 1892 
foregoing example, next, adding to the re- 478 
majnder one-fourth of 1912, also the number of 306 
days, 306, from January 1, up to and including — 4 


November 1; and 4 (19 + 4) and dividing by 7, 7)2680 
we get a remainder of 6; the sixth day of the 389.~6 Frid 

week, Friday, on which November 1 will happen; Shoat ish toute 
consequently election day will be the following Tuesday, or Nov. 5. 


Exam. 3. Abraham Lincoln was born in Kentucky, 


February 12th, 1809, on what day of the week did it 


happen ? 
1809 — 18 = 1781 


SoLuTion.— Deducting 18 from the given year 452 
leaves 1781; then proceeding according to the rule 43 
we find the remainder to be 1; the first day of the 4 
week, Sunday. 7) 2290 


On what day will it happen in 1912? Ans. Monday. 


176 CHRONOLOGICAL CALCULATIONS. 


Exam. 3. If your birthday be Dec. 29, 1900, on what 
day of the week will it happen ? 


1900 — 19 
1881 
475 
363 
4 


2728 


a 


9.... Saturday. 


Subtracting 19 from 1900, we get 1881. Adding the required 
numbers according to the rule and dividing by 7, we find the day to 
be Saturday, the division being exact. 


BISSEXTILE OR LEAP YEAR, 


Nore. — The solar year, or the time required by the earth to go once around 
the sun, is 365 da. 5 h. 48 min. 48 sec. The common year is 365 days. Hence, 


1 solar year is 5h. 48 min. 48 sec. longer than 1 common year. 
4 ‘* years are So As oe ode 4 = years. 
100 66 eG oe 24 da. 5 bb 90 G6 66 66 100 e 
400 Ge Ge 6s 96 66 21 66 20 &e 6% 6 400 be ee 


or 97 days nearly. Hence, 

If 97 days be added to every 400 years, in other words, if 97 leap years be 
reckoned in every 400 years, the calendar will be only 2h. 40 min. in advance of 
true time; or about | day in 4000 years. To be more explicit, let us take the 
time between the years 1600 and 2000, a period of 400 years. It is clear from the 
foregoing that we cannot reckon every fourth year in these 400 a leap year as 
that would give 100 leap years, while the correct number is 97, a difference of 3 
years. To distribute those 97 days, then, among 97 years, every fourth year in 
we 400 is reckoned a leap year, except the centennial years 1700, 1800 and 1900, 

ence, 

A fF Every year that is exactly divisible by 4 is a leap year, the centennial years 
excepted ; the other years are common years. 

Il. Ever y centennial year that is exactly divisible by 400 is a leap year ; ; the 
other centennial years are common years. 

The year 1900, for example, is a common year, because, although exactly 
divisible by 4, it is not exactly divisible by 400. The year 1904 isa leap year, be- 
ing exactly divisible by 4; and the years t600, 2000 and 2400 are leap years, being 
exactly divisible by 400. 


APPENDIX. 


INTEREST RULES TERSELY STATED AND EXPLAINED. 


GENERAL RuLe.— To jind the interest of a given sum for any 
number of days, at any rate per cent., on the basis of 360 days to 
the year: Multiply the principal by double the rate, and the 
product by the days, cut off five tigures from the right, counting 
Jrom the decimal point always, and add a third and a sixth. 


Exam. What is the interest of $3765 for 8 days at 3242 


$3765 
9125900 
758 
1255 
$3/1375 


Doubling the rate, 33, we get 74; multiplying this by 8, the 
days, gives 60 ; then 60 times the principal is $225900. Cut off 
five figures to the right, and add } and 1 of that third; the 
interest is $3.1375. 


Note.— When the principal is not large enough to cut off five figures, prefix 
a cipher or ciphers, to make five, and proceed according to the rule. Thus, if 
the principal were $65, in the example, 65 x 60=$3900; prefixing a cipher we 
have .03900, five decimal places, and adding a third and a sixth of that third 
we get .05446, the interest of $65 for 8 days at 3% 4%. Whenthere are cents in 
the principal there will, of course, be seven places of decimals. 


The reason of the rule will be understood from the following : 
The problem fully expressed is this : If $100 earn $33 in 360 
days, what will $3765 earn in 8 days? And the solution, by 
compound proportion, would be as follows : 


$100 : $8765 :: $382 
360 days : 8 days | 


Va = ee 


178 Interest Routes TERSELY STATED AND EXPLAINED. 


Reducing this to a simple proportion, we have the following: 
36000 : 30120 :: 32 


Doubling the rate, now, doubles the first term of the proportion, 
and we have 72000: 3u120 :: 7s. 

Multiplying now by 73, double the rate (we have already multi- 
plied by the days, 8), and dividing by 72000, solves the problem. 


Instead of dividing by 72000, in the usual man- 72000 
ner, we prefer to make use of a more simple divisor, 24000 
4000 


and that we get by adding a third and a sixth of 
that third, making 100000, as shown in the margin. 100000 


Bearing in mind now, that whatever operation is performed on 
the divisor to simplify the division, a similar operation must be 
performed on the dividend; hence we add to 225900 a third and 
a sixth of that third, having cut off five figures from the right 
which divides by 100000. 


Mernop or Exact INTEREST. 

GeneraL Ruie.— To find the interest of a gwen sum for any 
number of days, at any rate per cent., on the basis of 365 days to the 
year: Multiply the principal by double the rate, and the product 
by the days, cut off five figures from the right, counting from 
the decimal point, and add a third, a tenth and a tenth. 


Exam. What is the interest of $275000 for 50 days, 
at 2% ¢ 
$275000 


550(00000 
183/33333 
18/33333 
11833338 


$753/50000 
7530 


42 


Interest Rurtes Tersery STATED AND EXPLAINED. 17Y 


Doubling the rate, 2, gives 4; then 50 days multiplied by 4 
gives 200. Multiplying the principal, now, by 200, cutting off 
five figures from the right, and adding a third, a tenth of that 
third and a tenth of that tenth, we get $753.50. This gives an 
excess of 10 cents in every $1000 of the interest, or 1 cent in 
every $100, so we bring 10 times the dollars, or 7530, to the right, 
and subtracting, we get the correct interest, $753.42. 

The reason of the rule will be understood from the following: 
The problem fully expressed is this: If $2 be paid for the use of 
$100 for 365 days, how much will be paid for the use of $275000 
for 50 days, at the same rate? And the solution by compound 
proportion would be as follows: 


$100 : $275000 :: $2 
365 days : 50 days 
Reducing this to a simple proportion, we have the following: 
36500 : 13750000 :: 2 
Doubling the rate now, doubles the first term also, and we 
have: 
73000 : 13750000 :: 4 


Multiplying the middle term now by 4, and dividing the pro- 
duct by 73000 solves the problem. 


Instead of dividing by 73000, however, we prefer 73000 
to use a more simple divisor, and that we get by 243334 
adding a third, a tenth of that third and a tenth ° 24334 
of that tenth, making 100010, as shown in the 2434 
margin. 100010 — 


Performing a similar operation on the dividend 55000000, we 
add its third, a tenth of that third and a tenth of that tenth, 
getting 75350000 for new dividend, to correspond with 100010, 
the new divisor. Cutting off five figures divides by 100000 
(which, by the way, may be done at the beginning or end of the 
process), and rejecting 10 times the quotient, or the dollars, $753, 


180 Inrerest Rutes Tersety Sratep anD EXPLAINED. 


makes the required correction. (See example 3, page 53; also 
rule, page 54.) 


Nore.—If the principal be not large enough to give five decimal places, 
prefix a cipher, or ciphers, to make five, as in the rule for 360 days. 

If we had to use 366 days (leap year) as divisor, the complete divisor would 
then be the double of 366, or 732, multiplied by 100, or 73200, and by adding to 
this a third and a tenth of that third, the new divisor would be 100004, a 
simple divisor. (See example 4, page 54.) 


SPECIAL RULEs. 


Although the foregoing rules are general, there are special rules 
for certain rates which will be found shorter for business purposes. 
Foremost among these we would place the 


Six Per Cent. Meruop. 


Norrt.—This method has been already fully explained in the chapter 
on Interest, commencing at page 122, where we have made proportion and 
cancellation the groundwork of the rule, but for persons who may not have 
a knowledge of these subjects, the following will be found, perhaps, more 
simple and clear: 


Rue. Zo jind the interest of any sum for one year at any rate 


per cent.: Multiply the principal by the rate per cent. and divide 
the product by 100. 


Exam. What is the interest of $475 for one year at 6%? 


Multiplying by 6, the rate. and dividing by 100, we $475 
get $28.50, the interest. 28[50 


Suppose, now, we wished to find the interest of $475 for 2 
months, or 60 days, at 6%, on the basis of 360 days to the year, we 
simply take § of $28.50 (2 mos. or 60 days being ¢ of a year), which 
gives the required interest, $4.75, or a cent for every dollar of the 
principal, showing that we need never figure 2 mos., or 60 days’ 
int. at 6%, only call the dollars of the principal so many cents, in 
other words, 1% of the principal is always the interest for 60 
days at 64. 


InterEst Rutzes TrErRsELY STATED AND EXPLAINED. 181 


Taking this for our basis, it is evident that the int. of $475 
for 6 days at 6%, is a tentn of 60 days’ interest, $4.75, or ATS 
and for 3 days, the interest is half of 6 days’ interest, or  .2875 

Again the interest of $475 for 600 days, 20 mos., or 1 yr. 8 mos. 
at 67%, is 10 times the int. for 2 mos. or 60 days, $4.75, or $47.50 | 
and for 40 mos., or 3 yrs., 4mos., it is 2 times $47.50, or _ $95.00 

The rule will be found equally simple when the number of days 
is not a measure or a multiple of 60 or 6. 

Thus, if it were required to find the int. of $475 for 13 days at 6¢. 

Moving the decimal point two places to the left, gives the in- 
terest for 2 mos. or 60 days, and moving it three places, gives 
_ the interest for 6 days, at 6%, always. 


Now the interest of $475 for 6 days is .........202.42- 475 
raaor l2idays, iis 2 times: :475, Or Joie. . eels ile se .950 
Petts Peay itwisys Of) (ATD NOT, oc ys wees Geb wens 79 
Pate nerinverest £0T 19 GAYS OF &. soc soc ic dng eats dine _ $1.029 


And the interest of $475 for 117 days, at 6%, would be found 
thus : 


SMEeresGICOr (CAV 1S, aes, inc aw noes ctnescinacs $4/75 
And for 120 days, it is 2 times $4.75, or.......... oe AOU 
Pomardays, the int: is half of :475, or. ... 0... eee ewes 2375 


Which is deducted, leaving the int. for 117 days or.... $9\2625 
Exam. What is the interest of $420 for 279 days, at 6% ? 


Dividing 279 days by 60, it is contained 4 times, with a remain- 
der of 39. (See note 3, page 142.) 


- The interest of $420 for 60 days is........... wcccceee . $420 
And 4 times $4.20 is the int. for 240 days, or.......... 16,80 
The interest for 30 days is half 60 days’ int. or ......... 2,10 
And 8 times 21 cts. (the int. for 3 days, or {, of 30) is 9d. or 63 
Giving the interest of $420 for 279 days............... $19/53 


Now since the interest of $420 for 279 days is the same as the 
interest of $279 for 420 days, the foregoing example can be sim- 
-plified by taking the dollars for the days, thus: 


182 Interest Ruutes Tersety STratep AND EXPLAINED. 


The interest of $279 for 60 days, at 6%, is............4. $2|'79 
And for 420 days, it is 7 times 60 days’ interest, or...... $19/53 


(This method can be always adopted when it ismore convenient 
to take the dollars for the days, 


Nott.—It need scarcely be remarked that should the principal contain 
cents, it will not affect the process, the figures to the right of the line being 
simply decimals. 


Exam. What is the interest of $345.60 for 1 y. 10 mos. 
26 d., at 43% % 
Moving the decimal point one place to the left, gives the 

interest for 20 months, or for 1 y. 8 mos., at 6%........ $84/560 


The interest for 2 mos, is a tenth of that, or........... 3/456 
And for 20 days, the interest is a third of 2 mos., or . 1/152 
The interest for six days 18... .. .. 5. ..3s Sen 345 - 
Giving the interest of $345.60 for the given time, at 6%, 39|513 
Deducting } of the interest at 6%, $39.518, or .......... 9/878 
Gives the interest at 4397, or. . 22. ...0 ssc nse see $29 1635 


Nore.—The difference between 6% and 4147 is 114; and 1 is 4 of 6. (See 
page 124, also rule, page 125.) 
INVEREST ON RUNNING ACCOUNTS, 


Instead of finding the interest on each item separately, as is 
frequently the case with persons having to deal with such matters, 
it will be found much more simple and expeditious to proceed as 
follows: 


Exam. What is the interest of $3000 for 23 days; 
$4500 for 10 days; $5000 for 19 days; $2000 for 26 
days; and $4000 for 17 days, at 6% ? ? 

Multiply each principal by the number of days respectively; 


cut off three figures to the right, from the sum of the products; 
divide by 6, and the result is the interest at 62. 


InterEst Routes Trersety STATED AND EXPLAINED, 183 


Process, 
$3000 X 238 == 69000 
€4500 X 10 = 45000 
$5000 X 19 = 95000 
$2000 X 26 = 52000 
$4000 xX 17 = 68000 
329|000 
$54(833 
Note.— For any rate other than 6%, add or subtract the difference, as | 
pointed out at page 124. 


! 


Exam. Find the interest on the following at 62 : 


Process. 
$2700 for 3 y. 5 mos, 12 days. 8100 13500 32400 
$1500 for 5 y. 7 mos. 8 days. 7500 10500 12000 
$4500 for 7 y. 9 mos. 13 days. 31500 40500 58500 
$3800 for 1 y. 3 mos. 20 days. 3800 11400 76000 
$5000 for 2 y. 0 mo. 28 days. BOOGIE pa HAS ee 140000 
$7500 for4 y. 1 mo. 0O days. 30000 7500 


909.00 834.00 318.900 


$5454.00 | $417,00 $53.15 


Adding the three results thus found, we get the required 
interest, $5924.15. 


Rue I. Multiply the principal by the years, months and days, 
setting the product of each respectively in a separate column, as shown 
in the example, and add. 

IL. Point off two figures to the right in the sum of the yearly col- 
umn, multiply by the rate, and the result is the interest for the years 
at the given rate. 


184 Inrerest Ruutes Trersety STATED AND EXPLAINED. 


III. Point off two figures in the monthly column and take half ; 
and for the daily column point off three figures and divide by 6 ; the 
result in both cases will be the interest at 6% from which the interest 
at any other rate is readily found by addition or subtraction, as 
already pointed out. | 


NotEr.— Reason of the rule: The interest of $2700 for 3 years is the same 
as the interest of $8100 for 1 year; the interest of $2700 for 5 months, the same 
as $13500 for 1 mo.; and of $2700 for 12 days the same as $32400 for 1 day, at 
any rate per cent.; consequently the interest of the totals, $90900; $83400 and 
$318900, for 1 year, 1 mo.and 1 day respectively, is equal to the interest of the 
several principals for the given time and rate. 

Pointing off two figures from the right of the dollars in any Brincipat and 
multiplying by the rate, gives the interest of that principal for1 year at the 
given rate. 

Pointing off two figures from the right of the dollars in any principal, 
gives the interest for 2 mos. at 6%, and half of this is one month’s interest. 

Finally, pointing off three figures gives the interest for 6 days at 6%, and 
therefore the interest for 1 day is 1-6 of that. 


ImporTANT Facts to BE REMEMBERED. 


(1.) That the interest of any principal for 6000 days, at 6%, is equal to the princi- 
pal itself ; in other words, any sum of money will double itself at 6%, sim- 
ple interest, in 6000 days, 200 months, or 16% years. 

(2.) Moving the decimal point one place to the left in any principal, gives the 
interest of that principal for 600 days, at 62. 

(8.) Moving the point two places gives the interest for 60 days ; and 

(4.) Moving it three places gives the interest for 6 days, at 6%: thus: 


The interest of $4765 for 6000 days, at 6%, is $4765 


€ ‘* $4765 “ 600 days, ie $476.5, or $476.50 
a ‘© $4765 ** 60 days, OS BAGS 
Se ** $4765 “* 6 days, ri $4.765 


And from this simple basis the interest for any time and rate can be easily 
found, and often by a choice of two methods: (See exam. page 181), 


.ExaM. Find the interest of $2000 for 119 days, at 64. 


First Method. Second Method 
$20|00= 60 days’ interest. Take the days for the dollars, and the 
40\00=120  ** 3 Ee dollars for the days: 
Be chy oh rs Spat 6d. \ Int. of $119 for 6000 days is $119. 
BES E2000 os Nate or Sears 


‘Spar Pee 7 be 
$39/67=119 ‘“ [ int.=33e. | 


A SIMPLE METHOD FOR AVERAGING ACCOUNTS, 


In averaging, there are two kinds of equations, Simple 
and Compound. 

A Simple Equation has reference to one side of an 
account only, which may be either a debit or credit. 

A Compound Equation has reference to both sides of 
an account. 


SimpLeE EQuATION. 


If one person owe another, on Jan. 1, $300 payable in 
4 months, $500 payable in 6 months, and $400 payable 
in 103 months; at what time may the whole be paid 
without loss to either party ? 


Process. 


$300 x 4 — 1200 
500 x 6 —= 3000 
400 x 10% — 4200 


ale Sr 


The interest of $300 for 4 months equals the interest of $1200 
for 1 mo.; the interest of $500 for 6 mos. equals the interest of 
$3000 for 1.mo.; and that of $400 for 104 mos. equals the interest 
of $4200 for1 mo. And since the interest of $8400 for 1 month 
equals the interest of $1 for 8400 months, $1200 = 7 months. 


186 A Srmerte Meruop ror AvErRAGING ACCOUNTS. 


The time, therefore, is 7 months from Jan. 1, or Aug. 1. 


Rute. Multiply each payment by its term of credit, and divide the 
sum of the products by the sum of the payments; the quotient will 
be the time to be counted forward from the date at which the credits 
begin. . 


Exam. What is the equated time of payment for the 
following bill? 


New York, Jan. 1, 1893. 
EDWARD JONES. 


To JAMES FRENCH SOR. 

1892. June 5, To Cash .......<sscedsenese es ee $300 
July 20,‘ Mdse., 3-mos....).. 329i. cae 400 
om Rage 10;***o Mds6., 2 MOR sacs oe eee (ean eee 200 
ee Oct. 1, Cash ac. ss ewe 2» yin pectin oe 500 


Instead of proceeding school-boy fashion, arranging due dates, 
counting the days, etc., let us treat the time as months and the 
fractions of a month, as in the preceding example. 


Notr.—It need scarcely be remarked that, in averaging an account, we 
can assume the first date on the bill, or the last date, or any date outside 
the bill, as the date from which to reckon, or the focal date as it is called. 


Mo. 
Q June 5, $300 1500 400 
1 July 20,3 mos. 400 8000 1600 
2 Aug. 10, 2 mos. 200 2000 800 . 
45 -Oct. <1; 500 - 500 1200]0 2000 
$1400) 4800(8 mos. 
4200 
600 
18000(18 days 
18200 


Sept. 13th. 


A Simpete Meruop ror AvERAGgING Accounts. 187 


EXPLANATION.— Assuming for convenience, the last day of the 
mouth previous to the earliest date of the bill, or May 31, as the 
focal date, we see that the time on the first item is 5 days; on the 
second, 4 mos. 20. days; on the third, 4 mos, 10 days, and ou the 
last item it is 4 mos. and 1 day. 

The days or dates are fractions of a month, viz., 35, 29, 49, etc., 
and instead of multiplying by each fraction separately, we simply 
multiply by the numerators, or dates, add the products and divide 
by 30 (cut off one figure and divide by 3), which gives a monthly 
product. 

Dividing 12000 by 30, we get 400 which is carried to the right, 
as shown in the margin; next, we multiply by the months, adding 
those on the left-hand margin to these on the right of the dates, 
viz., 1 and 3, or 4; then 4 times $400, or 1600, is set under 400, 
to the right; now 4 (2 + 2) times 200, or 800, and finally 4 times 
500, or 2000. Then dividing the sum of the products, 4800, by 
the sum of the payments, 1400, we get 3 mos. and a remainder of 600, 

Multiplying this rem. by 30, and dividing the product, 18000, by 
1400, we get 13 days, nearly. The time, then, is 3 mos, and 13 days, 
counted forward from May 31, giving the average due date 
Sept. 13th. 


Norr.— The advantage of arranging the time in months on the margin, as 
shown in the example, wil] be apparent to the reader, as it enables us to see 
at a glance the whole time on each bill, including the term of credit. 


To make this simple method of average more clear, it may be well to 
remark that, had we multiplied $300 by #5 mos.; $400 by 422; $200 by 
429 and $500 by 435, and added the results, we would have got $4800, as 
found by the short process on opposite page. 

Again, since $300 for 5 days — $1500 for 1 day; $400 for 20 days = $8000 
for 1 day, &c.; therefore, $12000 for 1 day equals the several bills for their 
respective number of days. Dividing $12000, the amount for 1 day, by 30 
(days in a mo.), gives $400 for 1 mo., or the result for all the dates or frac- 
tions of a month. 

Multiplying each bill, now, by its respective number of months, setting 
the results under $400 and adding, we get $4800, or the result for 1 month. 
Now, if we have the use of $4800 for 1 mo., how long ought we to have 
the use of $1400, the amouni of bill? Ans. As often as $1400 is contained 
in $4800; the statement by proportion being; As $1400: $4800::1 mo. 
Dividing 4800, now, by 1400, we get 3 mos. and 13 days, which means that 
the interest of $4800 for 1 mo. = the interest of $1400 for 3 mos. and 13 days. 


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A Smrete Mernop ror Averacine Accounts. 189 


Find the equated time of payment, and the cash bal- 
ance of the foregoing account, Jan. lst, 1894, int. at 6%. 


Process. 
Mo: Dr. 293 
see an. 129 2 mos. 100 1200 200 
2 Mar. 12 1 mo. 300 3600 900 
6 July 20 3 mos. 200 4000 1800 
600 880\0 3193 
Mo. Cr. 212 
2 Mar. 10+38 2 mos. 150 1950 600 
5 June 26+3 38 mos. 100 2900 800 
9 Oct. 15 100 1500 900 
350 6350 9512 
=) ke mos. 
500 
181 
canbe d. 
5500 


Ans. March 22; Cash balance, $261.63. 

Arranging the time on both the debit and credit sides of the 
account as pointed out in the preceding example, we find the 
sum of the payments on the Dr. side 600, and the sum of the 
products, 3193. 

And on the Cr. side the sum of the payments is 350, and the 
sum of the products, 2512. 

Taking the two latter sums from the former, we find the bal- 
ance of acct. 250, and the difference of the products, 681. Divid- 
ing the latter by the former, we get the time 2 mos. 22 d., and 
counting forward we find the average date to be March 22d. 

Now the time from March 22, 1893, to Jan. 1, 1894, is 9 months 
and 9 days, and the interest of $250, the balance of acct. for that 
time, at 6%, is $11.63, making the balance due, $261.63. 


Norte 1. If the balance of the acct. be not contained in the difference of the 
products, multiply the latter by 30, then divide and the result will be days. 

Norte 2. It may be well to remark that, in case of a note or draft, the days 
of grace are to be taken into consideration, as in the two first items of the 
credit side where we have multiplied by 13 and 29 days. 


‘2681 ‘eT “400 oJep osBIOA oY} Suryeu ‘Og judy ‘eyep [vo0j oY} WI} punmyong peyunod eq Of, 


‘p LI ‘SOU 9 = OOF + 089% 
0892 -007$ 
LOTS 00ST 
SELP O|o09T o008T$ LOTS OlOTTS  o09Ts 


eee ee — Se 


000T 00sg = 000 xX 63 “PO & 009 ooe9 = 008 x Te Ame 

00FS 0099 = 009 X IL 3des oF 008 O0sZt = 008 X gL eune ft 

008 0092 = 007 X 6 Alne g S0L 000g =o0ss Xp API 0 

SEG “aO iba ‘ud ron 
; "880004 


00 00% pliataliole gusie'ts (6) 6107400 sale, ©. oe yseg Rs 6Z "190 00 008 ee ee ee 5 ee IZ Ane 
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°190 A Simpte Meruop ror AVERAGING ACCOUNTS. 


A Simpte Meruop ror AVERAGING Accounts. 191 


Here, we observe, that although there is a balance of $400 on 
the Dr. side, the Cr. side has an excess of $2630 for 1 month; in 
other words, J. B. has had the advantage of that amount for 1 
month more than his creditor; hence, he should pay the balance, 
$400, soon enough to make the terms of credit equal. 

Now, since $1 = 26380 mo. 

$400 = 6 mos. 17 days; 
that is, the interest of $1 for 2630 months, or $2630 for 1 month 
equals the interest of $400 for 6 mos. 17 d. 


Hence the following 


Ru.Le.— I. Arrange the time in months, on the margin, taking the 
last day of the month previous to the earliest date of the account, as 
the focal date. 

Il. Multiply each payment by its respective date, add the several 
products, cut off one figure, and divide by 8. 

Ill. Under the result, set the products of the payments by the 
months, and take the difference between the debit and credit products, 

IV. Divide this difference by the balance of account ; the quotient 
will be the average term of credit, to be reckoned FORWARD from the 
focal date when the balance of products and balance of account are on 
the same side of the account ; but BacKWARD when the balances are 
on opposite sides. | 

V. If the balance of account becomes due before the date of settle- 
ment, add the interest for the interval, at the legal rate; if after, 
deduct the interest. 


MONTHLY STATEMENT. 


Notr.—It is scarcely necessary to remark that, in averaging an account, the 
cents in the payments are not multiplied, the rule being to reject them when 
less than 50, and when 50 cents, or more, to add $1. 


EXAMPLE. 

1894, Jan. 2. To Mdse.. $885 770 8 
Berea rt tees eee 270 1080 12 
(Ohh. th ee pa 2520 24. 
ihc gee Gre mk bah SO: 560 38920 42 
lta Ui Mima h ech ie bee tee 8 3450 30 
Be tS ina, ORs aA 4944 48 
hinds BSE RN fala Peter de a, 7800 75 
CSA CS fae ee eae COS rhein 600 10800 108 
A DU eer ee 480 9600 - 100 
Oo ee ener e tice, 160 4000 50 
ay 1 FIM NUR i 40 1120 00 
pint: (RS ch Ueda At) 180 5400 60 
SEC BLE rite: Goths ue 200 6200 62 

$4572) 61604(13 46)619(138 


Multiplying each payment by its respective date, and adding, 
we find the sum of the payments to be $4572, and that of the 
products, 61604. Dividing the latter by the former, we get Jan. 
18th the average date. 

The same result, observe, is obtained by simply multiplying the 
dollars by the date, omitting the two right-hand figures of the 
dollars, taking what is nearest the true result always, thus: Call 
$385, 4 and multiply by 2 (that is, $400 x 2), $270, $300, or 
3 xX 4; 4 X 6, etc., and divide by 4600, that is, 46. 


Nots.— Count from, and include, the first day of the month always. 


A Smee MetHop ror AVERAGING Accounts. 193 


Average the following account. Terms, 60 days: — 


1895. 
Mo 
0 April 3. To mdse. $16 48 
Laas & 25 175 
Gs 8 120 39 
_1 May 6. 12 72 
eg 10 70 
«19, Gass ete é4 
“25, ry 175 
8 July 8. 15) 120 
<< 12. 20> 41 240 123 
Ese 16, 6) 96 
$124 8|0)117/6 124)196(1 mo. 
— 124 
39 —— 
72 
30 
124)2160(17 
2108 


Arranging the time in months on the margin, we find 0 mo. for 
April, 1 mo for May and 8 mos, for July 

Multiplying each item of the bill now by its respective date, and 
adding the results, we get $1176 for 1 day. Dividing this by 30 
(days in a mo.) gives $39 for 1 mo., which is carried to the right, 
opposite the last date of April. Then $34, the amount for May, is 
multiplied by 1 mo. and $41, for July, is multiplied by 3 and.the 
results set under $39 and added, making $196 for 1 mo. Dividing 
196 now by $124, the amount of bill, we get 1 mo. and 17 days. 
Counting 1 mo. and 17 days forward from April 1, inclusive, we get 
May 17th for the average date; and counting 60 days (term of 
credit), forward from May 17, gives July 16 for the due date. (See 
note, page 187. 

To impress more thoroughly upon the mind of the student, a 
knowledge of this important subject, it may be well to give the solu- 
tion of one more example, as follows: 


194 A SimpLe Meruop ror AVERAGING ACCOUNTS. 


John King wishes to settle his account on Dee. 1, 
1895; how much does he owe, charging interest at the 
rate of 6%? 


1895. Dr. 
Mo. 
Q May 7. Tomdse. $25 175 
aes Reh ak aioe 30 270 
nS LE A Seana 20 300 82 
Oy VULY SL ene eee 5 9) 
Meta Uae oe ey oF 280 210 
meat Bet be TO 60 720 
BOcte Vega Se 8 24 
en Uh ea | $72 520 360 
(e ibaat Ler 12 )- 180 
$252 3)0)247\4 $652 
82 
Cr. 
1 June12, By Cash. ae $70 600 45 
cine Fae mle ia had 20 300 70 
B COct, ctoestai 100 100 
Teale yeas rorcnsal aieare aaa a 
00 80) BT 
$52 B&B 52)113/2 mo. 
104 
a 
30 


52)270(5 days. 
260 


SOLUTION.— Counting May 0, July 2, Oct. 5 on the Dr. side, and 
June 1 and Oct. 5 on the Cr., we have the time arranged at sight on 
the margin. Proceeding now according to the rule, we get $652 for 
1 mo. on the Dr. and $765 on the Cr. side. The balance of account 
is $52, and is on the Dr. side; the balance of interest is $113, and is 
on the Cr. side. In other words, the balances are on the opposite 
sides of the account, aud therefore the time, 2 mos. and 5 days — 
(found by dividing the balance of interest by the balance of account), 
is to be reckoned backward from May 1, giving Feb. 238d for the 
average date. Interest is now charged on $52 from Feb. 23d to Dec, 
1 (date of settlement), at 6%. The time is 9 mos. and 6 days; the 
\nterest, $2.39, making the balance due, $54.39. : 


SHORT METHODS. 


Exam. What is the cost of 1278456 pounds of iron at 
$24.812 per gross ton (2240 Ibs.) ? 


Lone METHOD. SHort METHOD. 
1278456 12784516 
24.812 18263 657 
639228 9131/828 
319614 ’ 4565|914 
278456 456/591 
10227648 5|'707 
5113824 2/853 
2556912 1 426 
cae alate Le ai 322 $14164/322 
2240 
9328 Dividing the number of pounds 
8960 by 70, gives the price at $32.00 
a 1 of $32“ ‘“ 16.00 
14408 aor t ties Bee a 8. 00 
13440 zy of $8 us ns . 80 
9681 atthise ie oot 01 
8960 4 of this a4 cé 4 
7217 ; = 5 
4978 Adding we get price at $24.81 
4480 (See exam. 1, page 109.) 
4980 
4480 


et 


Note.— The division by 2240 can always be simplified by cutting off one 
figure from the right of the dividend, counting from the decimal point, and 
taking a quarter, one-seventh, and one eighth. thus: 


196 ‘Suort Muruops. 


Cutting off one figure and dividing by 4, 

een a divides by 40; then one-seventh of that 

1133141577 quarter, and one-eighth of that seventh, be- 

"$14164|329_ cause these numbers are factors of 2240 (40 
. x7 x 8 = 2240). 


Exam. What is the cost of 259356 pounds of iron at 
$13.75 per gross ton % 


25935)6 
3705|085 
926|271 
463/136 
115/784 
57/892 
238/946 
$1592/029 
Cutting off one figure and dividing by %, gives the 
PPICE Abo. LEM wicis seve ew ee © 5 208 90 'e ene $32. 
One-fourth of $32.18 the price at... ... 4c... eee 8. 
One-half of $8 * EMP Sr es ee 4, 
One-fourth of $4 $$ §Oe i Saw de «eee arerpe Ie 
One-half of $1 os SS Sta ae ee ee 3:0 ole eee .50 
And one-half of this RE Bese © 8 0.9-a\e\e eieneeee 25 
The total.sum is the price at.......*..sss«es evens $13. 75 
Or thus: 
Cutting off one figure, and dividing 925935]6 
by 8, divides by 80, and gives the 3241195 
price at $28 per gross ton always. 16201975.. $931 .568 


Half of $28 is the price at $14, or 281946 
25c. more than the given price. To $1592 029 
get the price at 25c. we set one-seventh 
of the price at $14, or $231. 568, a little to the right of the work, 
as shown in the margin; this is the price at $2, and an eighth of 
this, or $28. 946, is the price at 25c. Subtracting this fram the 
price at $14, gives the price at $13.75. 


Suort Mernops. 197 


Notst.— From the foregoing examples, it will be readily seen how easily 
the cost of any number of pounds, at $14, 16, 21, 24, 25, 28, 32, 34, 36, or in fact 
any price per gross ton, may be found. (See note to exam. 3, page 111.) 


Exam. What is the cost of 562800 pounds of iron at 
$37.50; of coal, at $3.75; and freight at 3874c¢. per 
gross ton ¢ 


| IRON. CoAL. FREIGHT. 
56280/0 5628/00 $562|800 
8040|00 —$32.00]| 804/00 —$3.20 80/400 = .32 
1005|00 = 4. 100j00°== 40 10/(050== 4 
goli25 = 1 LoL eos CLO Aiol2=—= 1 
125|625— .50 12/562—= 5 1/256 = cota 


————————|/ |__| —. SS eee 


$9421|875—= $37 .50|| $942|\187— $3.75 $94/218 = .374 


Cutting off one figure and dividing by 7, gives the price at $32 
per gross ton always; cutting off two, and dividing by 7, gives 
the price at $3.20, or a tenth of $32; and cutting off three 
figures, and dividing by 7, gives the price at 32 cents, or the one- 
hundredth part of $32; and the aliquot parts are the same. 

Or, we could have proceeded in either case as if the price were 
$37.50 per ton, and take one-tenth or one-hundredth at the finish. 


Oats. 


Exam. What is the cost of 4760 pounds of oats at 
593,c. per bushel (82 lbs.) ? 


47/160 = .32e. 
23/80 — .16 
1190 — 8 
D975 == 2 
14487— 1 1 
185:=== fr 
3 — ts 


198 SHort Meruops. 


Assuming the price at 1c. per pound, we have the price at © 
once, at 2c. per bushel; then 16c. equals 4 of 32; 8c. half of 16; — 
2c. a quarter of 8; 1 a half of 2; 2, or its equal, 3, is an eighth - 
of 1c.; and 54 is half of that. 


Exam. What is the cost of 300 bags of oats, 79 pounds 
net to the bag, at 25fc. per bushel ? 
$237|00 — 32¢. 


118|50 —16 
59/95 — 8 
7/406 —= 1 
3I703— 4 
1s51— 2 

995 — 4 


$191|635 — 252 
300 multiplied by 79, gives the number of pounds, 23700, and 
at 32c. per bushel, the price is $237.00. Half that is the price at 
16c., or $118.50; at 8c. the price is half that, or $59.25, and 1c. 
is one-eighth of that, or $7.406; §, or its equal, 4, is half of 1c. 
or $3.703; 2 is half that, or $1.851, and 4 is half that, or .925, 
making $191.635, the price at 2d{c. 


Corn. 


Wheat, buckwheat, barley, etc., may be treated in like manner, 
by assuming the price at a cent per pound, or at as many cents 
per bushel as there are pounds to the bushel. 


Exam. What is the cost of 2940 pounds of corn at 676. 
per bushel (56 Ibs.) ? 


$29/40 — 56c. 
4/20 = 8 
1105 — 2 
sone] 


$35/18— 670, 


SHort Meruops. 199 


Cutting off two figures we have the price at 56c. per bushel ; 
one-seventh of which is the price at 8c.; 2c. is a quarter of 8; 
and ic. is half of 2, giving $35.18, the price at 67c. 


And if the quantity should be given in bushels and pounds, the 
same method of solution can be employed, as illustrated in the 
following: 


Exam. What is the cost of 364 bu. 27 lbs. of oats at 
428c. per bushel ? 


364/27 


eer 


29/12 


116)75 — 32a 
291187 — 8 
7|296 — 

91 oa 
456 — 


8154/1601 498 


| | 
bo 
jm oojao 


Multiplying the bushels by 8, and setting the result two places 
to the right, we get 2912; multiplying this, in turn, by 4, and 
adding at the same time the 27 lbs., we get 11675 lbs., or the 
number in 36423 bu. 

The remainder of the process needs no further explanation. 


Notr.—To reduce pounds to bushels, we should never make use of long 
division when the number of pounds to the bushel is a composite number. 
Thus, how many bushels in 11675 lbs. of oats ? 


11675 
2918 ....3 
364....27 


200 SHort Mertuops. 


Simply use the factors, 4 and 8 (4 X 8 — 82). Dividing first — 
by 4, we get 2918, and a remainder of 3. Next, we divide by 8, 
and we get 364 bu. and a rem. of 6; this rem. is multiplied by 
the first divisor, 4, and the rem. 3 added, making 27 lbs. 

How many bushels in 14763 lbs. of corn ? 


14763 
2109 
263....35 


Here the factors are 7 and 8 (7 x 8 =56). Dividing by 7, we 
get 2109; dividing this by 8, we get 263 bu. and a rem. of 5, 
which is multiplied by 7, giving 35 lbs., or the complete rem. 


Hay. 


Exam. What is the cost of 1860 pounds of hay ¢ at 
$17.50 per ton (2000 lbs.) ? 
18/60 
2/325 
$16|275 
At 1c. per pound it is $18.60, which is $20 per ton. The dif- 


ference between $17.50, the given price, and $20 is $2.50 which 
is 4 of $20. Deducting 4 we have the price at $17.50. 


- Pounps to Gross Tons. 


To reduce pounds to gross tons we have the following simple 

Rule: Cut off one figure from the right of the pounds, and take a 
quarter, a seventh and an eighth; the result will be gross tons and BS 
decimal of a ton. 


Exam. How many gross tons in 79952 lbs. of iron ? 


SoL.— Cutting off one figure, and taking a quar- 7995/2 
ter, 4 of that quarter, and { of that seventh, we 1998 8 
285|5428 


have 35.6928+ gross tons, (See exam. page 196. —35 16098 


STHRLING. 


Pounps, SHILLINGS AND PENCE. 


Rue. Jo reduce shillings, pence, etc., to the decimal of a pound : 
Divide the pence ana farthings (having reduced the farthings to a 
decimal) by 12; to the result thus found prefix the shillings, and 
divide by 20. 


Exam. 1. Reduce 18s. 22d. to the decimal of a pound. 


Reducing $ to a decimal we have 2% =........ ar tale 2.75 

mividing this’ by 12,:we have: ..i.0...5....5 wath swe .22916' 
Prefixing the 18s. to this result, we have ............ 18 .22916' 
mruing now Dy. 20, we Nave; i...» cies deinceise alec + £.9114583’ 


Note.—To divide by 20, move the decimal point one place to the left and take 
the half. 


Exam. 2. Reduce 19s. 103d. to the decimal of a pound. 


Seeeretial. , AHO LOR 1S CQUAL ow. weiss ss ares Coe eas 6% 10.5 
Dividing this by 12 and prefixing 19, we get........... 19.875 
mre TA LE FOS G WC TOb eo ace ars)s «oases ce 'ae,e o's s £.993875 


Reverse Rote. 


To find the value of the decimal of a pound steriing to the nearest 
farthing: (1) Take % of the number expressed by the jirst two figures 
of the decimal for the shillings of the result. 

(2) Diminish the number expressed by the remainder, with the third 
Jigure of the decimal annewed, by sz of itself, what remains will be the 
Sarthings in the rest of the required value. 


202 STERLING. 


Exam. 1. What is the value of £.99375. 


Taking a fifth of 99, the first two figures of the ,99375 
decimal, we get 19 shillings with a remainder of 4, 19s. 104d. 
to which 3 (the third fig.) is annexed, making 48, 
this is diminished by 34, leaving 42 farthings or 
104 d. 


Reason: .99875 is more nearly equal .994 than .993, and 
.994 = .95 + .044, and the value of .95 is found by multiplying 
by 20 and dividing by 100, or simply ¢ of 95 = 19s. Then the 
value of £.044 or £;44, by diminishing the denominator by s we 
have 960 (farthings in a pound sterling), and by diminishing the 
numerator 44 by »~, we get 42 nearly, hence £44, is nearly equal 
£& #2, or 42 farthings, 


Exam. 2. What is the value of £.8525. 


Taking 4 of 85 we get 17s. and the third figure 1%s. 04d. 
is 2 farthings, that is 3 or 4. 


To Divine Pounps, Suinuines AnD PrncE By 100. 


To divide pounds, shillings and pence by 100, in other words, 


to take 1% and consequently any per cent. of sterling money, we 
give the following simple 


RuLe.— Hor the pounds of the quotient take the pounds of the 
dividend, except the last two figures, which are to be divided by 5 for 
shillings ; from the remainder, with half the shillings annexed, reject 
sz part and regard what remains as farthings. 


Exam. Take 1% of £8947..138..8d. 
£89.. 9s. .64d. 


STERLING. £03 


Here by cutting off 47, the last two figures of the pounds, we 
have £89; and + of 47 is 9, the shillings required, and the re- 
mainder is 2. This remainder with 7, the half of 14s. (because 
13s. Sd. is more nearly 14s.) annexed becomes 27, from which 
1 is rejected (nearly its 3), we have 26 farthings or 64d.; the 
answer is £89 9s. 64d. 


Note.—In rejecting the 1-23 always take what is nearest the true result. 
The reason of the process will be understood from the preceding rule. 


Exam. What is 4% of £89..16s..4d.? 


Dividing 89 by 5, we get 17, the shillings, 
and a remainder of 4 to which 8 (the half of £89 16s. 4d. 
16) is annexed, making 48 farthings, from 
this 2 (nearly s,) is rejected, leaving 46 far- 17s. 114d.. 
things, or 114d. This is 1¢ of the given sum. Lost ls: LOG: 
Then 4 times this is £8 11s. 10d., or 4% of the 


given sum. 


Reason of the rule: Dividing £89 by 100, we get £0, leaving a 
remainder of £.89. If this be multiplied now by 20 (shillings in a 
pound), and the result divided by 100, we get 17s. But multiplying 
by 20 and dividing by 100 is the same as dividing by 5; so we simply 
take a fifth of £.89 for the 17 shillings of the answer. This leaves 
a remainder of £;4,, or £.04. Now 16 shillings — £46 or £,8,, writ- 
ten decimally, £.8 is still to be divided by 100; this gives £.008. 
We have now £.04, the remainder left in dividing 89 by 5, plus 
£.008, which make, when added, £.048, or, when expressed frac- 
tionally, £;43,. Diminishing the denominator of this fraction, now, 
by ss part of itself, we have 960; and diminishing the numerator 
also by a similar part of itself, we have 46 nearly ; hence, £,4%, is 
nearly equal to £,4°,, or 46 farthings, since £5351 farthing. 


204 STERLING. 


STERLING REDUCED To AMERICAN CURRENCY. 


The first three figures of the decimal of a pound will be sufficient 
for all practical purposes in reducing Sterling to American currency; 
and to find those three figures, the following simple rule is given, 
which will be found preferable, perhaps, to bee given for finding 
the decimal in full, at page 201. 

Rue. —(1.) Take half the number of shillings for the first ‘a 
of the decimal, tf the shillings be even ; and if odd, half the shillings 
will be the first two figures of the decimal. (2.) Reduce the pence and 
farthings, if any, to farthings, by multiplying by 4, and vf the result 
consists of only one figure, set tt in the third decimal place, but if of 
two figures, set them in the second and third places, adding 1 ¢f the 
number of farthings iz between 12 and 36, and 2, if between 36 and 
48. (3.) Prefix the pounds to the decimal thus found, and multiply 
by the rate of exchange. 


Exam. What must be paid in New York for a bill on 
London for £32 18s. 23d., the rate of exchange being 
P4.874 ? 


SoLutTion.— The shillings being even, half OPERATION. 
the number, or .9, is the first figure of the deci- O48 GIT 
mal. There are 11 farthings in 23d., and this 4.875 
number forms the second and third figures of $160 .441125 
the decimal. We have now £32 18s. 2?d.— 
£32.911, and this multiplied by $4.875 = $160.44. 


Notre.— The multiplication by the rate of exchange can be frequently 


shortened if we assume $5 asa standard rate. In that case, at $1 per pound 
sterling, the cost of £32 18s. 234d. or its equal, £32.911, would be $32.911; at $2, 
two times that, and at $5 it would be five times $32.911 or $164.555. 

Now multiplying by 5 is the same as multiplying by 10 and taking half, so 
that if we move the decimal point in £32.911 one place to the right, thus, 
£329.11, and take the half, we get $164.55, or the value of £32.911 sterling, im 
American currency, at $5 per pound sterling. 

And since the difference between $5 and $4.8714 is SHORT METHOD 


124% c. or % of a dollar, we a the value at $4.873¢ by 329/11 
subtracting x of 82 91Lor A. of 164.55 (4 of $1 being 164/55 
the same as 3 |, of $5). sia a 


a ae ~—— 


STERLING. 205 


Again, since the value of £32.911 at $1 per pound sterling, is $32.911, the 
value at 10 c. would be the tenth of that at $1, or $3.2911, and at 1 c. the value 
would be a tenth of the latter, or $.32911; at 2c. it would be twice $.32911; 
at 3% c. three and a half times $.32911, etc.; so that, had the rate of exchange 
been $4.88, $4.89¥, or $4.91, we would simply add to $160.44 (the value at 
$4.8734), once $.32911, or 33 c.; twice $.32911, or three and a half times $.32911 ; 
1,2and 3¥% being the differences between $4.8734 and the three mentioned 
rates. If the rates were less than $4.8744, we would, of course, deduct 
the difference. 


To illustrate further this simple method let us take 
the following: 


Exam. What is the value of £612 17s. 9d. at $4.86 2 


SoLuTION. — Taking half the 17s. decimally, we have  .85 
for the first two figures; 4 times 9d.— 36 farthings and 1 make 37 


to be set in the second and third places, making ... .. Ret PeOOk 
the three decimal figures required. We have now £612.887 to be 
multiplied by the rate, $4.86} which gives the required value, in 
American currency. 


SHorT METHOD, 
Moving the point in £612.887, or, which is the 6128/87 


same in effect, drawing the vertical line one 80641435 _ 
place to the right, and taking half, we have ie ak 
$3064.435, the value at $5. Then 7, of this S064 
(simply a fourth set in proper position) is $76.61, 11539 


which being deducted from $8064.435, would — WH |——"_ 
give the value at $4.874. But the rate is $4.86, $2983|229 

or a difference of } of a cent, which is also to be deducted, 3 =} 
and +; half of $6.128 (value at lc.) is $3.064, and half of this is 
$1.532, the value at 1 of a cent. Taking these three results now 
from $3064.435, we have $29838.229, the value at $4.86}. 


Nortr.—In subtracting the three items from $3064.435, the process is per- 
formed by addition,-thus 2 and 4 are 6, and 9 (setting down 9 at the bottom), 
are 15 (the top figure); carry 1 to 3; 4 and 6 are 10, and 1, are 11 and 2 at 
bottom, are 13, etc. 


506 STERLING. 


Or we might proceed as follows: Suppose the rate — 


had been $4.883, what is the value of £612 17s. 9d.? 
SoLution.— The difference between $5 (our standard rate) and 

$4.883, is 113 cents. 

£612 17s. 9d.= £612.887, and at $1 per £1,the value= $612.887 


And at $5 per pound aie the Values 2, ce.e ea 
At 10c. per £1, the value is ;4, of that at $1.. = 61/288 
And at 1c., the value is +5 of int at 100,27 seen 6/128 
At éc., or its equal, tc., the value is half 1le.. ... = 3/064 
And {ce. is a quarter of $, or } of Ic..... eer 766 

$2993] 189 


These four items make up the value at 113c., and deducting them 
from $5, or rather from its value, $3064.435 (by the method of addi- 


ee ea ee a a ae ee 


tion, as pointed out in the foregoing note), we get $2993.189, the © : 


value at $4.882. 


The reason for taking half the shillings, etc., to find the first three figures 
of the decimal, will be understood from the following: 
16s. =£48, 55%, ifs, or -3, all representing the same value differently 

expressed: therefore 17s. = £47 or 785)... 0.2.2... wee eee ae 
And 9d. reduced to farthings =£,25,; now if we add to the de- 

nominator 950, +1 =z Of itself, or 40, we get 1000, and by increas- 

ing the numerator 86, by a similar part of itself, we make it 37 

nearly ; so that £35, is nearly =£335 .-: sees eeees o eye sey See 


Consequently lis. 9d. expressed decimally...... ........ .... = .887 


In reducing 9d. we simply say 4 times 9 are 386, and 1 are 37; and lis to be 
always added when the number of farthings is nearest to 24, or when the 
number is between 12 and 36, as stated in the rule. 


InrerEst on STERLING. 
Rute I. To find the interest of any sum for 1 year, at any rate 
per cent: Multiply the principal by the rate, and divide by 100. 
Exam. 1. What is the interest of £1 sterling for 1 year 
at 5% per annum 4 


a ee ee 


STERLING. 207 


SoLtution.— Multiplying £1 by 5, the rate, and dividing by 100, 
we have £;3,, or £.05, for the interest. 

But £35 = £55, or 1 shilling, or its equal, 12 pence; and since 
the interest of £1 for 1 year, or 12 months, at 54%, is 12 pence, the 
interest of £1 for 1 month is one-twelfth of 12 pence or 1 penny, 
and consequently the interest for 2 mos. is 2 pence; 3 mos. 3 pence, 
etc. Hence the following: 

Rue Il. To find the interest of any number of pounds sterling 
for a given number of months, at 5% per annum: Take the pounds 
as pence and multiply by the months. 


Exam. 2. What is the interest of £42 for 7 mos. at 5%? 


SoLuTion.— Calling 42 pounds 42 pence, or 3s. 6d. and multiply- 
ing by 7, the number of months, we get £1 4s. 6d. the interest. 

So likewise the interest of £42.. 10s. for 8 months at 52%, is 424 
pence, or 3s. 63d. X 8= £1.. 8s. 4d. | 

And from this the interest at any other rate may be easily derived. 
Thus, to find the interest at 4%, simply subtract from £1 8s. 4d. one- 
fifth of itself, or 5s. 8d., and we have £1 2s. 8d., the interest of £42 
10s. for 8 mos. at 4%. Had the rate been 6% we should have added 
one-fifth of the interest at 5%; or we might employ the following: 


Rue Ill. To find the interest for months at 6% per annum: 
Multiply the principal by half the number of months, and divide the 
result by 100. 


Exam. 1. What is the interest of £439 16s. 8d. for 1 
year and 8 months, at 6% per annum ? 


OPERATION. 
SoLutTrion.— 1 y. 8 mos.= 20 mos. £439 16s. 8cde 
Multiplying £439 16s. 8d. by 10, half of 20 mos., 10 
we get £4398 6s, 8d., and dividing this by 100, - £4398 6s. Sd. 
we get £43 19s. 8d., the interest required. £43 19s. 8d. 


To divide by 100, we set down £43, leaving a remainder of 98. 
Taking one-fifth of 98, we get 19s. and a remainder of 3. To this 
remainder we annex half of 6s., or 3, making 33 from which & is 
rejected, leaving 32 farthings, or 8i., nearly. (For short method 
of dividng by 100, see rule page 202; also reason of rule page 208.) 


208 STERLING. 


Exam. 2. What is the interest of £756 14s. 10d. for 
5 months, at 54% per annum. 


This example is solved by both the 5% and 6¢ rules as follows ; 


First Method, 5 per cent rule. 

SOLUTION. — £756... 14s. 10d... ese eee 7563d. nearly 
Multiplying this by 5, the number of mos. we have 37833d. 
Dividing this by 12 (pence in a shiliing) we have.. d1ds. 33d. 
Dividing the latter by 20 (shillings in £1) we have £15 1ds. 33d. 

the interest at 5%. Now there are 20 quarters, or 

A4ths in 5%; therefore 1 quarter, or 32 is 54, of 52. 


Dividing £15 15s. 38d. by 20 we have............. 15s. 94d. 


Which is added making the required interest..... £16 11s. 1d. 


Second Method, 6 per cent rule. 
Multiplying £756 14s. 10d. by 5, we have ........ £38783 14s. 2d. 
And dividing this result by 2, we have........... £1891 17s. 1d. 
This multiplies by 24, or by half of 5 mos..... 
Dividing this by 100 (short method), we have..... £18 18s. 44d. 
Which is the interest at 62. 
The difference between 6% and 54 is 3%. 
There are 24 quarters in 6%; therefore 3 is 4 of 24 

quarters,-or of 6%. Taking 4 of £18 18s. 44d., 

we get the interest at 2%....... -...ceedscorees = £2 Ws) Brae 


Which is taken from the interest at 6%, leaving ... £16 11s. 1d. 
the required interest. 


Rute tv. To find the interest of any sum for any number of days 
at any rate per cent: (1.) Reduce the shillings and pence, if any, to 
a decimal by the rule given at page 204. (2.) Multiply the principal 
by double the rate, and the result by the number of days, cut off five 
figures from the right, counting always from the decimal point, and 
add «a third, a tenth of that third and a tenth of that tenth. 


Nots.— In computing interest on sterling for days, the basis is 365 days to 
the year. 


STERLING. 209 


Exam. 1. What is the interest of £648 15s. 3d. from 
June 2 to Nov. 25, at 54 per annum ? 
SoLutron.— There are 176 days from June 2 to Nov. 25, 


Reducing 15s. 8d. to a decimal, we have the principal = £648.762 
Multiplying this by 10, double the rate, then by 176, 


Ne CR lass A layne ek Fifi oes os «ens 2 vba sce £11/41821.12 
Cutting off five owes from the ee and taking} 
ee et Gate Wie iy Gi oe oe ae eaye'e Ge vis-e oh scene ne 3|80607 
Taking a tenth of this third, we get.......c.csecee 38060 
And taking a tenth of this tenth, we get ........... ° 3806 
Adding the four results now, we get .........s..e06 ~ £15 64294 
150 
Rejecting 10 times 15, or 150 from .64294, we get.... 64144 
_ Finally, the value of £.64144 is 12s. 10d. nearly, 12s. 10d 


making the required interest £15 12s. 10d. (See rule, page 201.) 


Exam. 2. What is the interest of £8000 for 75 days 
at 4% 2 


SOLUTION.— £8000 K 600... ..-. ccc ccc eceee = £48/00000 
Multiplying the principal by 8, and by 75, or by 600 16 
(75 X 8), at once, we get £4800000. Cutting off five 116 
places to the right, and adding a third, a tenth of 16 
that third and a tenth of that tenth, we get £65.76000. —@pelsannn 
Rejecting 10 times 65, or 650 from this, we have £65)76000 
£65.7535. Then taking one-fifth of .75, we get 15s., 650 
and the third figure of the decimal now is simply 7535 
3d., making the interest £65.. 15s. 03d. lds. 02d. 


(See General Itule, page 178 ; also reason of the rule, page 179.) 


OruER SHort Meruops. 


The six per cent rule of interest already explained in this work, 
can be applied to the computation of other matters, such as, Wheat, 
Clover Seed, Potatoes, etc., where the number of pounds to the 
bushel is 60. 


210 ; Orner SHorr. MEtTuHops. 


Exam. What is the interest of $1860 for 1 y. 9m. 5d. 
at 6%? 


SoLuTion.— In 1 y. 9mo. 5d., there are 635 days. 


The interest of $1860 for 600 days....... Se = $186)0 
Half $18.60 (60 days’ int.) is the int. for 830 days... = 9/380 
One-sixth of $9.30 (80 days’ int.) is 5 days.... ... = 1/55 
Making the interest of $1860 for 685 days... .... = Bi0G|s8 


Or, taking the dollars for the days and the days for 
the dollars, we have the interest of $635 for 1860 days, 
thus: 


SoLuTIon.— The interest of $635 for 600 days. = $63)5 
Three times 600 days’ int. is the int. for 1800 days. = ae 
One-tenth of the top line, or $63.50 is 60 days.... = 6/35 
And the two results, when added, is the int. for —_——/|— 

1860 days os... acs cwat, es hies una er os $196/85 


(See example, page 181). 

Suppose now we change the foregoing problem into the 
Sollowing : 

Exam. What is the price of 1860 lbs. of clover seed, 
at $6.35 per bushel (60 lbs.) ¢ 


SoLuTron.— At 1c. per lb., or at 60c. per bushel, the price of © 
1860 lbs. would be 1860c. or $18.60. 


And the price of 1860 lbs. at $6, or 600c. per bushel= $186|0 
At 30c. per bushel, it is half the price at 60c. or 

halt $19/60.0.0 00 wna ee ee = 9/30 
And at 5c. the price is one- an of 30, or of $9.30..= 155 
Making the price of 1860 lbs. at $6.35 ..... pears = $196 85 


The same as was found by the first solution of the interest problem. 
And uf the problem be reversed so as to read: 685 lbs. of clover seed. 
at $18.€0, or 1860c. per bushel, it can be solved same as the second 
solution of the interest problem. The student will please try it. 


OtTxeER SHort MeErsops. 911 


W HEAT. 


Exam. What is the cost of 16940 lbs. of wheat at 
87ic. per bushel (60 lbs.) ? 


SOLUTION. — Cutting off two figures from 16940 gives the cost 


at 1c. per pound, or at 60c. per bushel... ....... = $169\40 

One-third of the cost at 60c. gives the cost at at 20c.= 56I47 

At 6c. the cost is a tenth of the top line, or of 60c...= 16/94 

At 1c. the cost is a sixth of that at 6c.............. = 9/82 

And at 4c. it is the eighth part of 1c .............. = 35 

Making the total cost i,0¢iii les es $245 198 
Porartogs. 


Exam. What is the cost of 749 lbs. of potatoes at 474c. 
per bushel (60 Ibs.) ? 


Sou.— Cost at 60c. .. 


II 


$7.49 | Or thus: 


Tan eee rs 3 >, | Cost at 6c. per bushel.= _—$.749 

trans 3 15K, — 1.87 | And 8 times 6c.=48c..= $5.992 
¢ of PCa aac...) — dl | Less, of .749= jJc.= 62 

ee Alize ...= $0.93 | And the cost at 47ic.= $5.93 


Hints on Inrersst. 

Savings banks allow interest on deposits for a certain 
fixed term, generally 3 mos. or 6 mos., and calculations 
may be simplified in many cases by multiplying the time 
and rate together and dvniding the result by i, which 


_ will give the rate for the given time, thus: 


“ee 


Exam. What is the interest of $872 for 3 mos..at 4% 
per annum ? 


Sovurion. —*<* = 1%, and 1% of $872 = $8.72, the interest; the 


process being, of bins performed mentally . 


912 | Hints on Interest. 


And if the rate were 34% per annum, we would deduct one-eighth 
of itself from $8.72, and the remainder is the interest for 3 mos. 
at 342. 


Exam. What is the interest of $648 for 3 mos. at 334 
per annum ? 


SoLuTion.— The interest at 4% is simply 1% of $648.... = $6.48 
And at 4%, the difference between 47% and 34%, itis} of4%. = .81 


Making the interest for 3 mos. at 8334 per annum........ $5.67 

Again, the interest for 3 mos. at 2% per annum, is half that at 4¢; 
at 5¢ it is 4% plus + and at 8% it is twice 4%, etc. And the same rule 
will be found to hold good in many other cases. it 


Exam. What is the interest of $648 for 1 year and 5 
mos. at 7% per annnm ? 


SoLUTION.— 1 y. 5 mo.=17 mo.; 9147-102 less +. 
The interest of $648, at 17 = $6.48, and at 102 it is... $64.80 
From this we deduct 54, of 1% or of $6.48, equal...... , 4 
Leaving the interest for 1 y. 5 mos., at 7% per annum, $64.26 

And so with other problems of a similar nature.. 


Montaty Payments, 6%. 


When interest is to be calculated on monthly payments, 
we have the following simple | 

RULE. (1) Add 1 to the number of months. (2) Multiply the sum by 
half the number of months, and the result by the payment, and divide 
by 2. (8) Lf the payment be in dollars, point off two decimal places, 
and if there are cents, point of four places. 

Exam. Suppose a person to pay into a Building and 
Loan Association, $3 a month for 15 months; what is the 
amount due at the end of the time at 6% interest ? 


Hints on INTEREST. 213 


SoLUTION.— 15 + 1 = 16; then mSEesS, or OES? $1. 80 
And adding the principal, $3 K 15 ............. a 45.00 
We have the amount, that is, the prin. and int.... = $46.80 
And for 


WEEKLY Payments 6%. 
We have the following 


Route, (1) Add 1 to the number of weeks. (2) Multiply the sum by 
half the number of weeks, by 7 and by the weekly payment, and divide 
by 6. (8) If the payment be in dollars, point off three decimal places, 
if there are cents, five places. 


Exam. Suppose a person to pay in 40c. a week for 10 
weeks; what is the amount due at the end of the time at 
6% interest ? 


SoLuTion. —10-+ 1=11; then -2*47*-.... = & 032 
And adding the principal, 50c. x 10........ a RE 5.00 
Dweenave tho AMOUNt........sser-eee es Mee eee we ees $5 .032 


And from this the interest at any rate other than 6% can be found 
by aliquot parts, as has been already pointed out. 


Reason of the rule: If the payments were $1 per week, and the problem 
solved by the usual method, we sliould have multiplied $1 by 10; $1 by 9; $1 
by 8, and so on down to the last payment; and adding the results, we get 
$55, the principal for 1 week. But we see that by adding 1 to 10, and multi- 
plying by 5, we get $55 more readily; and 7 times $55 or $385 is the principal 
for 1 day, at $1 payments. Now the payment is 50c.; multiplying $385 by .50, 
we get $192.50, the principal for 1 day at 50c. payments. And the interest of 
$192.50 for 1 day at 6%,is found by multiplying by 1 day, pointing off three 
figures from the decimal point, and dividing by 6, and we have .1920+6= 
.03208 +, or .032, as shown in the example, for the interest. 

The same line of reasoning is applicable to the monthly rule. Multiplying 
$3 by 15, 14, 13,12, and so on down tothe last payment, and adding the results: 
we get $360, the principal for 1 month. But this is more readily found by 
adding 1to15, and multiplying by 7¥, or, which is the saine in effect, 15 by 
8, the half of 16; then by 3, and we have $360. The interest of $360 for 2 mos. 
at 6% is simply 1% or $3.60; and for 1 month it is half, or $1.80, hence the 
reason for dividing by 2. 


914 Hints on INTEREST. 


A SmpLe Mernop. 
To find the face of a note, the proceeds being given: 


Rup. (1) Find the interest of the proceeds for the given time and 
vate. (2) Find the interest of that interest, and so on, till the interest 
ts so small as not to affect the result. (3) Add the interest thus found 
to the proceeds, and the sum is the face of the note. — 


Exam. For what sum must a note be drawn at 3 mos. 
to net $2500 when discounted at 6%. 


SoLUTION.— $25.00 is the interest for 2 mos. on........ $2500.00 
12.50 . ‘s- ToTO; 
Making. . . $37.50 a << 3 mos. 
The int. of $37.50 for 2 mos. = 88c. 
“ 1 m6. 19¢. 
Making the int. for. 3 mos. = 5%7c.; and the whole int.= 38.07 
Adding $38.07 interest to the proceeds, we get.......... $2538 . 07 


the face of the note. 


Norte.— The days of grace being abolished in the State of New York, are 
not taken into account. If grace be allowed, add the interest for the days 
of grace. : 


Proof: The interest, or bank discount, for’ mos. at6Zon $2538.07 
(the sum for which the note is drawn), is found to be 08.07 


and deducting this from $2538.07, we have the net proceeds. $2500. 
Exam. 2. Having discounted a note in bank, I am 


credited with $1500 as the proceeds, what was the face 
_ of the note, the rate being 4% and the time 90 days? 


SOLUTION. —The interest for 90 days, or 8 mos. at 4% on $1500 
is 1% or $15; and on $15, the int. is 15c. making... .... 15.15 


and adding this tu the proceeds, we have the face....... $1515.15 
Proof: The interest of $1515.15 for 3 mos. at 4% is 1% or 15.15 


and deducting this from the face, we have the proceeds.. $1500.00 


Pt oe 
Ee 


Hints oN INTEREST. — a We) 


From the foregoing hints, we derive the following 
simple rule for computing 


InreREstT FoR Monrtus. 


Rute. 70 find the interest for months at any rate per cent: Mul- 
tiply the principal by the product of the months and rate, and divide 
by 12. Or, multiply by the months, then by the rate, and take haif 
the result, and one-sixth of that half. Or, multiply the months and 
rate together and divide by 12; the result is the rate for the given 
time ; multiply the principal by this rate. 


Exam. What is the interest of $480 for 5 mos. at 5%? 


480 x 25 $480 K5 x 5 
SoLuTIon.— BAS 2 $10.; Or, — = $10. 


Or, 5 times 5 = 25, divide by 12 = 257; then $480 x 2, = $10. 

Or, because $480, the principal, is divisible by 12, we have 40 x 
25 = $10, the required interest. 

This rule, after a little practice, is so simple that, in numerous 
cases it will not be necessary to use pencil or paper in computing 
interest, thus: What is the interest of $72 for 8 mos. at 344? 

Ans. 8 times 3} is 28; then 28 multiplied by 6 (42) = $1.68 the 
interest. Or, 28 + 12= 23, and $72 x 24% = $1.68; and if there be 


YEARS AND MoNTHS 


Reduce the years to months and proceed according to the rule. 


Note. —In conclusion, it may be well to remark that, in relation to the 
foregoing rule, the rate may be changed for the time and the time for the 
rate, and very often to great advantage; thus, the interest of any sum for3 
mos. at 4% is the same as for 4 mos at 3%; for 6 mos. at 5%, the same as for 
5 mos. at 6%; for 3 mos. ana 15 days at 3%, the same as for 3 mos. at 336%, 
8 mos. 15 days being equal to 344 mos.; etc. 

Suppose it were required to find the interest of $320 for 3 mos. and 15 days, 
at 3%. Tosolve this, we reverse the problem so astoread: $320 for 3 mos. 
at 34%. 

Then the interest for 3 mos. at 4% on $320, is simply 1%.. ........... $3.20 
and at %% itis an eighth of 4%, or 40c. deducted.................. 40 


giving the interest for 3 mos. and 15 days at 3%........... Siscacesdne: €o\ Oas0U 


INTEREST SIMPLIFIED. 


If 360 days be divided by the rate per cent, the interest of any sum 
of money for the number of days thus found, at the given rate, will be 
equal to one per cent of the principal. 


Thus if the rate be 44%; dividing 360 by 44 gives 80 days; now the 
interest of any sum, say $1768 for 80 days, at 444, is found to be 1% 
of that sum, or $17.68; and for 8 days the interest is one-tenth of the 
latter, or $1.768; for 800 days, the interest is ten times $17.68, or 
$176.8, and for 8000 days it is ten times the last interest, or $1768. 
In other words, the interest of any principal for 8000 days, at 442, is 
100 per cent, or equal to the given principal. 


Hence we have at sight: 
The interest of $1768 for 8000 days, at 447=$1768. 


ce 6é 800 66 cé = 176.8 
66 ce 80 ce ce = 17.68 
¢é “cc 8 €é ce = 1.768 
And the interst for 1 day: i! Serer 


And from this simple basis the interest for any number of days is 
easily found. An example will make this clear. 


Exam. What is the interest of $3765.47 for 92 days 
at 44%? 
Here we have at sight, $37.6547 80 days’ interest; 


Next, we have 3.76547= 8 “ 
And half of this, or 1.88273= 4 < 


Making $43.3029 —92 « 


IntTEREST SIMPLIFIED. 217 


From the foregoing illustrations and examples the following will 
be readily understood : 


“Rate. 360 days. 
4% 90 ‘ hence, 9000, 900, 90, 9and 1 day. 
44g 80 ** Sram SOUO ABOU; 80,78 Foal 
Bigewee sos | SES TZ00, TAU A725 Bees 
6% ee a ee GO00 7600760,-6 <<! 
74% 260648 —OSS <¢ 4800, 480, 48,8 < 
Beer) db pis! «¢ 4500, 450, 45, 5 * 
et GO Ei ‘« 4000, 400, 40,4 “ 

10 % Dox nS «¢ 3600, 360, 36,6 <“ 
Lesa. 730) ** *« 8000, 300, 30, 3 “ 


fhe ep 
nw 
nw 


T he interest of any principal, at any rate, for the number of days 
placed under 360, opposite the rate, will always be one per cent of the 
given principal. 


Suppose now it were required to find the interest of $2460 for 93 
days at 5%. (93 days=72+18+3). 


Here we have at sight, $24.60 =72 days’ int. 
Next, we have one-fourth of this, or 6.15 =18 < 
And then one-sixth of this or TO 5 < 


Making $31.775-938  « 


Nort.— If the rate be 514%, add a tenth of the interest at 5%, 34% being one- 
tenth of 5%; and if 534%, add for the 14% half the interest at 14%, etc. 


To make this simple and interesting method of interest more clear, 
let us take another 


Exam. What is the interest of $1872 for 94 deys, at 
3% 4 
First, 94 days=80+12+2. 


By referring to the table we see that 74 is contained 48 times in 
360 days. We have now, at sight, the interest of $1872 for 4800 
days, 480 days and 48 days, viz. $1872, $187.2—$187.20 and $18.72 


918 IntEREST SIMPLIFIED. 


respectively; and from this, by a slight mental effort, we have the 
interest for 24 days, 16 days, 12 days, 8 days, 6 days, 4 days, 3 days, 
2 days and 1 day. | 


Here then, we have the interest for 480 days =$187 .20 
And for 80 days, the interest is 1 of $187.20 = $31.20 
12 days’ int.=} of 48 days, cr fof $18.72= 4.68 

2 44 =fof 12 “ ,or¢tof 4.68= ite: 

Making 94 days’ intercst, at 742 = $36.66 


And if the rate were 72, we would add ,}, of the interest at 747° 
and if 742%, deduct 55; + being #5 of 747, or 30 quarters, 

To divide by 30, move the decimal point one place. to the left, or 
suppose it moved, and divide by 3. 

Notr.—It is scarcely necessary to remark that the interest of any sum for 90 
days, at 4%, equals the interest for 45 days at 8%; in other words, when the rate 
is doubled the days are halved, and vice versa. 

If, now, we refer to the example given at page 177, under General Rule, we 
find that the problem there given, can be more readily solved from a knowledge 
of what has been given in the table. 

Now the rate given in the problem referred to‘is 3%, which is half 744%, and 
since 48 is the number of days which will give 1% of the principal for interest 
at 736%, the number of days for 334 is 96. Or 360 days+334=96 days. Hence 

The interest of $3765 for 96 days, al 3342=$37 .65 
and 8 ‘* =), = 3.1375 


Importanr Facts Inuusrratep By EXAMPLEs. 


If we bear in mind the fact that the interest of $1 for 2 days, at 
any rate per cent. is the same as the interest of $2 for 1 day; $60 
for 189 days, the same as $189 for 60 days ; $5000 for 1 y. 7 mo., or 
570 days, the same as $570 for 5000 days, etc.; many problems of 
daily occurrence in business transactions can be simplified. 


Exam. What is the interest of $8000 for 5 mo., 19 da., 
or 169 da. at 249? 


Dividing 360 days by 24, the rate, we get 160 days as the basis for 
that rate; in other words, the interest of any principal for 160 days, 
at 24%, is one per cent. of the principal always. 


INTEREST SIMPLIFIED. 919 


The interest, therefore, of $8000 for 160 days, at 24% = $80.00 


:s u Bees dr = 4.00 
Lass 4 = 97.50 
Making the int. for 5 mc. 19da. or 169 « = $84.50 


But the problem can be simplified at once by reversing it so as to 
read: $169 for 8000 days, at 24%, as in this case we have the interest 
at sight by simply taking half $169=$84.50 

REASON .— Since the interest of $169 for 160 days at 2147, is $1 69, the interest 
for 1600 days is ten times that, or $16.90, and for 16000 days it is ten times $16.90, 
or $169; in other words, the interest of any principal at 214%, simple interest, 
will equal the principal in 16000 days, and for 8000 days the interest is half the 
principal. 

Or, we might reason thus: It has been shown at page 216 that any sum of 
money for 8000 days, at 4147, simple interest, will double itself, or the interest 
will equal the principal, and at 2\44¢ it is half that at 44¢¢. And from this the 
interest of any part or any multiple of $8000 can be easily derived. 


To CuHancEe CommerciAL InterEst To Exact INTEREST. 


RuLE. Move the decimal point in the commercial interest two places 
to the left ; under the result set a third of itself, one tenth of that third 
and a tenth of that tenth; add the four numbers and take the total 
_ from the commercial interest, the remainder is the exact interest. 


Exam. What is the exact interest of $7200 for 147 
days at 5%% 


Reversing the problem we have $147 for 


7200 days; and at 5% the commercial interest = $147. 
Moving the point in this two places $1.47 
to the left, and adding a third, a tenth 49 
anda tenth, we get $2.0139, which is taken 49 
from $147, giving the exact interest 49 $2.0189 
$144.9861. $144. 9861 


Reason.— Commercial interest exceeds exact interest by aos, or ae part, 
and the foregoing is a simple method for dividing by 73. (See exam. 1, page 
51, and rule page 52.) 


220 INTEREST SIMPLIFIED. 


To Cuance Exact Interest to CommerciaL INTEREST. 


RULE. Move the decimal point in the exact interest two places to the 
left; under the result set one-third of itself and a sixth of that third; 
add the sum of the three numbers to the exact interest and we have 
the commercial interest. 


Exam. What is the commercial interest of a sum of 
money whose exact interest is $144.99 ? 


Moving the decimal point in $144.99 two places $1.4499 
to the left, and adding to this a third of itself 4833 
and a sixth of that third, we get $2.0137, which is 805 
added to the exact interest $144.99 giving the $2.0137 
commercial interest $147, very nearly. 144.99 

$147.0037 


Reason.— Exact interest is 30, or as part less than commercial interest, 
and to divide by 72 we add toit its third and a sixth of that third, which 72 
gives 100 fora simple divisor, as shown 1n the margin, making the same 24 
additions to $1.4499 to equalize; and moving the point two places to the 4 
left divides by 100. “106 


IntTEREST ON Datuy BALANCES. 


Ruue. Multiply each daily balance by the number of days it remains 
unchanged, and add the several products , the result will represent the 
principal for 1 day, Then if 365 days be taken as the basis, apply 
the Generai Rule given at page 178, and if 360 days, apply the rule 
given at page 177. 


Exam. What is the interest’ of $136800 for 1 day at 
2% (365 days) ? 

$136800 
Multiplying by 4, double the rate; cutting off five figures, 5.47200 
and adding a third, one tenth of that third, and one tenth 1.82400 
of that tenth, we have the required interest $7.49. 18240 
(See rule, and reason, page 178.) 1824 
$7 .49664 


Bank BAuancegs. O91: 


Balance. Deposit. Check. Balance. 
507-48 
$8396.16 $738.31 $492.52 $8641.95 


RULE. To find the final balance: Add the complement of the check 
Jigures to the deposit and original balance. Drop 1 immediately to 
the left of the last, or left-hand figure of the check, always. 


The complement of a number is what it wants of being a unit of 
the next higher order. Thus, 3 is the complement of 7 (7+3=10) 
and 48 is that of 52 (52+48=—100). The complement of a number is 
readily found by setting down, first, what the unit figure of that 
number wants of being 10; and next, what each succeeding figure 
wants of being 9. Thus, in the check number above, 8 is set down 
above the unit figure 2, so as to make 10; and next, 4, 7,0, 5, so as 
to make 9’s with the remaining figures. The complement of 
$492.52, then, is $507.48, the sum of both numbers making $1000. 

Instead of subtracting $492.52, now, to find the final balance 
$8641.95, we add its complement $507.48. The sum of the three 
numbers is $9641.95; but dropping 1 immediately to the left of the 
check figure 4, the true result is $8641.95. 


Reason of dropping 1: Subtracting a number from a unit of the 
next higher order, is the same as adding its complement and drop- 
ping that unit. 


Thus, taking $492.52 from $1000, is the $1000 $1000 
same as adding $507.48, and dropping the 492.52 507.48 
1 in 1000. } ———_  —- ——-—— 


$507.48  (1)507.48 


Nore. —In practice, the complement is not set down, the work being per- 

formed mentally. Thus, From the sum of $8396.16 and $738.31; take $492.52. 
. $8396.16 + $738.31 — $492.52 =$8641.95 

Here, we say § and 1 are 9, and 6 are 15; 5 and carry 1; 1 and 4 are 5, and 3 
are 8, and 1 are 9; 7 and 8 are 15, and 6 are 21; 1 and carry 2; 2 and 3 are 5, and 
9are 14; 4.and carry 1; 1 and 5are 6, and 7 are 13, and 8 are 16; 6 and carry 1; 
1 and 8 are 9; but here, 1 is dropped immediately to the left of 4, in the subtrac 
tive number, and, therefore, 8 is set down; the result is $8641.95. 

(For a fuller explanation on this matter, see page 283.) . 


Sy) Pig—Iron. 


In computing the cost of pig iron, 28 pounds are usually allowed 
for sand, making 2268 pounds to the gross ton, instead of 2240; and 
calculations are found to be tedious in such cases when made by the 
regular methods. 


The following rule will be found simple and practical where 2268 
pounds to the ton are used: 


RuLE. Hirst take one-seventh of the number of pounds, then one- 
ninth of that seventh ; the result will be the cost, vu dollars and cents, 


at $36 per ton. From this the cost at any given price is readily found. 


by aliquot parts, as in the example given at page 195. 


Exam. 1. What is the cost of 1728640 pounds of pig- 
iron at $12 per ton (2268 lbs.) ? 


Here we take } of the number of pounds, then 1728640 
4 of that seventh; the result is $27488.73, the 246948 .57 
cost at $36. Now the cost at $12 is 4 of the 27438 . 73 =$36 
cost at $36. =$9146.24—$12 


ReEAson: Instead of using 2268 as divisor, we prefer to make use of the factors, 
4,7,9 and 9 (4x7 x9 x9=2268) (It is immaterial in what order the factors are 
taken in the multiplication to produce 2268.) If we take 7x9 we have 63, leaving 
4x9, or 386. Hence we have 63x 36=2268. Now if we divide 2268 by 63, or by 7 
and 9 in succession, we get 36. 


Cancellation will often enable us to shorten the process, ag in the 
following: 


Exam. 2. What is the cost of 2742643 pounds of iron 
$14.40 per ton (2268 lbs.) ? 


Here we set the factors of 2268 on the 4% 274264/3 $14 48 
left, and the price $14.40 on the right, of 7 1097057 20 %.Q 
the pounds. We eliminate 4, on the left, 9  156722/457 .40 
and dividing $14.40 by 4 also, we get $3.60. YW $17413/606 
Next we eliminate 9, on the left, and divid- 
ing $3.60 by 9 also, we get 40c. Multiplying now by .40 we have 
$1097057.20, and taking 4+ of that, and 4 of that seventh, we get 
$17413.606, the required cost. 


, D 
Pe ee eee 


6 


Picg—Iron. 993 


Notr.— To solve this problem according to the rule, take 4; i of that seventh, 
and + of that ninth, and you have the cost at $12. Then $2is % of $12 and 40c 
is + of $2. Or, $2.40 is # of $12. 


And if there be tons and the fraction of a ton proceed as in the two 
following examples: 


Exam. 3. What is the cost of 403842 tons of pig-iron 
at $14 per ton ? 
“? 1872 $i 
Here we find the cost, first of i872 lbs. to be K =: 986 
$11.55, by the method of cancellation as in the 9 104 


preceding example. 9 $11.55 
Next, we have 40 14= $560 the cost of 40 tons =$560. 
which is added to $11.55, making total cost. = $571.55 


Another example will make this simple method clear. 


Exam. 4. What is the cost of 254988 tons of iron at 
$13.50 per ton ? 


Here we first find the cost of 1986 lbs. at $12, 1986 


as in example 1, by taking 4, $ and 4; or 4, 4 662 
and +, getting $10.508. 94.571 
Then $1.50 is 4 of $12, so we take an eighth $10. 508=$12 
of $10.508 to get cost at $1.50 — Pm es Veh 
Next, we have $13.50 25, or $1350--4 =$337 . 50 
Making the total cost = $349 . 321 


And if the price per ton were $13.75, we would simply add a 
sixth of $1.50, 25c being 4 of $1.50 


Nots.— It is scarcely necessary to remark that multiplying $13.50 by 25, is the 
same as multiplying by 100 and dividing by 4; 25 being 14 of 100. . 


924 Pic—Iron. 
REMARKS. 


The careful student need scarcely be told, at this stage of the 
work, that, in computing the cost of iron, etc., in quantities of 
pounds at so much per ton, much needless labor can be saved by 
reducing the pounds to tons and the decimal of a ton, instead of to 
tons and pounds as is most frequently done. 


To Renucr Pounps to Tons or 2268 Les. 


Rule. — Take one-fourth of the number of pounds, one-seventh of 
that fourth, one-ninth of that seventh and one-ninth of that ninth ; 
the last result is the number of tons and the decimal of a ton, and 
will represent the cost, in dollars and cents, at $1 per ton always ; 
and if British money, at £1. sterling per ton, ete. 


Note.— The reason for making use of these divisions is explained on page 222. 


Exam. What is the cost of 58668 pounds of iron at 
$10 per ton (2268 lbs.) ? 


Taking a quarter of the pounds, we get............. . 14667 


mne-seventh of this quarter = 2095.2857 
gne-ninth of the latter = 232.8095 
and one-niuth of the last es 25.8677 


Here we have 25.8677 tons, ex pressed decimally, instead 
of 25 tons 1968 Ibs. found by dividing by 2268. 


At $1 per ton, the ‘cost of 20.8677 tons 1S; 27a. eee $25.8677 


and at $10 per ton, the cost is ten times that = $258.677 


Notrge.—Should it be necessary to express the cargo on the invoice, or on the 
books, in tons and pounds it can be easily done, but the computations will be 
simplified by the method shown above, or better perhaps, iv the majority of 
cases, by the rule given on page 222. 


Se ee ee 


Pic—Iron. 225 
To Repuce Pounps to Gross Tons (2240 Lzs.) 


Rule. — Cut off one figure from the right of the pounds and take 
one-fourth, one-seventh of that fourth and one-eighth of that seventh ; 
the last result will be the tons and the decimal of a ton, and will 
represent the cost, in dollars and cents, at $1 per ton, always ; and if 
British money, at £1. sterling per ton, etc. 

Reason.—Because 40, 7 and 8, are component factors of 2240 (40x 7 x 8=2240). 


Cutting off one figure and dividing by 4 divides by 40; then 4 of that and 5 of the 
seventh, completes the division by 2240. 


Exam. What is the freight on 169840 lbs. of mdse. at 
90c. per gross ton (2240 lbs.) ? 


Cutting off the right-hand figure 16984/0 


and taking the parts as pointed out 4246 

by the rule, we get 75.8214 tons, 606/5714 
carried to four decimal places; and at $1 per ton = $75/8214 
At 10c. the cost is the tenth of that at $1 =e 10a E 


Taking this from the cost at $1, we get the cost at 90c. = $68.2393 


Suppose, now, it were required to find the freight on 347 cwt. 3 
qrs. 21 lbs. of mdse., in American currency, at 23s. 8d, British, per 
gross ton (2240 lbs.), the rate being $4.374. 


In 347 cwt. 3 qrs. 21 Ibs. there are 38969 lbs. 3896/9 lbs. 
Dividing now according to the rule, 974/225 
we get 17.3968 tons, carried to four 139|175 
decimal places; and at £1. or 20s. perton the cost = £17|3968 
3s. 4d. being 1 of £1. or 20s. we take 4 = 2/8994 

4d. is a tenth of 3s. 4d. (40d.) is pa 2899 
making the cost at £1. 3s. 8d., or 23s. 8d. =  £20/5861 

At $1 per pound sterling we have $20.5861; at $10 = $205.861 
Taking half $205.861, the value at $10, we get......$102.9305 
the value at $5. From this we take 4 of $20.5861 = 2.5732 
or 7, of $102.9305, and we have the required freight = $100.3573 


(See rule, exam. and note, page 204 ; also rules and reason page 201.) 


226 TonNnaAGE. 


To Repuce Gross Tons to Nat. 


RULE. — Add to the gross one-tenth of itself, and one Sith of that 
tenth. 


Exam. 1, How many net tons in 39 gross tons ? 


_ Adding to the gross one-tenth of 39 

itself, or 38.9, and one-fifth of 3.9 
this tenth, or .78, we get the net 78 
tons and the decimal of a ton eae 


43.68 = 43338. or 48 tons, 1860 lbs. 


Reason. To reduce gross tons to net, the grossis 2000 39 2000 
multiplied by 2240 and the result divided by 2000. 206 
Now, by taking 2000 for multiplier and 2000 for 4( 
divisor, and eliminating both, the multiplication 
by 2240 is completed by adding a tenth of 2000 and a fifth of that 
tenth; 200 being a tenth of 2000, and 40 a fifth of 200. 


Exam. 2. How many net tons in 47 tons 15 ewt. gross ¢ 


Since there are 20 cwt. ina ton, 15 cwt.= 5 or .75 of 47.75 
a ton; therefore 47 tons, 15 cwt.= 47.25 gross tons and by 4.775 
adding to this one-tenth of itself and one-fifth of that 955 
tenth, we obtain 53.48 net tons, or 53 tons 960 lbs. (.48 x 53.48 
2000 = 960.) 


Exam. 8. In 56 tons, 1860 lbs. gross, how many net 
tons ? 


56 
In this, we find first the net tons in 56 gross as in the 5.6 
foregoing example, by adding a tenth and a fifth of that 1.12 
tenth and we obtain 62.72 net tons. “$2.79 
Next we add to this $889 of a ton, which is .93 and we 93 
have 63.65 tons net, or 63 tons, 1300 lbs. cis 
63.65 
Exam. 4. How many net tons in 480 gross ? 
Adding one-tenth, and one-fifth of that one-tenth, we on ; 
obtain 537.6 net tons, or 537 tons, 1200 lbs. 9.6 


Note.—To reduce .6 of a ton to pounds, multiply by 2000. 537.6 


a ee 


TonNAGE. 997 


To Repucre Net Tons to Gross. 


Rue. — From the net deduct one-seventh of itself, and to the result 
add one-fourth of that seventh. 


Exam. How many gross tons in 43 tons 1360 lbs. net? 


43 tons, 1360 Ibs. or 435369 tons=..... aieiecd ae sc ae OS 
Deducting one-seventh of this, or 6.24 
6.24, we get 37.44 to which is 37.44 
added one-fourth of 6.24, or 1.56 1.56 
getting 39 gross tons. 39.00 


e 


Reason. —To reduce net to gross we multiply the net by 2000 and 
divide by 2240. 


By deducting one-seventh from 2240 43.68 2000 
each of the two first terms 320 6.24 
the relation of these numbers 1920 37.44 
is not changed. Nor is the 80 1.56 


relation changed by adding to 2000 39.00 

the remainders one-fourth of that 

seventh ; and the process equalizes the first and third terms. In 
other words, we have now to multiply by 2000 and divide by 2000, 
which does not affect 39. 


Or, to be more explicit, the numbers 2240 and 43.68 bear to each 
other the relation of divisor and dividend respectively, and if any 
change be made in the divisor a similar change must be made in the 
dividend to preserve the relationship. (See Gen. Prin. page 9.) 

Nortr. — It need scarcely be remarked that 112 and 100 can be used instead of 
2240 and 2000; and also that this method can be made use of in dividing any 


number by 112 or 2240: first performing the process as above and then dividing 
py 100 in one case and by 2000 in the other, at the finish. 


228 TonNAGE. 


Tue Net Cost Bretnc Given to Finp THE Gross. 


RULE. — Add to the net cost one-tenth of itself and one-fifth of that 
tenth. 


Exam. If a net ton of coal cost $4.50, what ought a 
gross ton cost at the same rate? 


Adding one-tenth and one- $4.50 
fifth of that tenth, we get the 45 
gross cost, $5.04. 9 

$5 .04 


Reason. — Since the cost of a gross ton is more than that of a net, 
at the same rate, we multiply the net cost by 112 and divide by 100, 
making use of the short method explained on page 226. 


THE Gross Cost Berne Given To Finp tHE Net. 
RULE. — From the gross cost deduct one-seventh of itself, and to the 
result add one-fourth of that seventh. 


Exam. If a gross ton of coal cost $5.04, what ought a 
net ton cost, at the same rate ? | 


Deducting one-seventh, and adding $5.04 
to the result one-fourth of that seventh 72 
we get the net cost, $4.50. $4.32 

.18 
$4.50 


Reason. —The reason will be understood from that given on 
page 227. It is simply a short method of multipiying by 2000 and 
dividing by 2240. 


' 
: 
3 
4 
7 
a 


TonNAGE. 229 


Net Tons. 


In computing the cost of merchandise in pounds, at so much per ton 
net, operations are simplified by assuming 1c per pound, or $20 per 
ton, as a standard price always. 


For example, the cost of 23760 pounds 


of mdse. at 1c per lb. or $20 per ton, net ............ = $237.60 
And at $2 per ton, the cost is a tenth of $20.......... = 238.760 
‘AG 20c.the cost.is a tenth of that at. $2...........0... = 2.8760 
And at 2c per ton, it isa tenth of 20c............ ede wee LO 


In other words, by moving the decimal point two places to the 
left in any number of pounds we have the cost at $20 per ton net, 
always; moving the point three places to the left gives the cost at 
$2 per ton; moving the point four places, gives the cost at 20c. and 
moving the point five places, gives the cost at 2c per ton. 

And from this simple basis, the cost at any given price may be 
readily obtained. 


Exam. What is the cost of 23760 lbs. of soft coal at 
$2.50 per ton net ? 


At $2, the cost is $23. '760 Or thus: 
And at 50c itis} of $2 = 5.94 At $20, the cost = $237.60 


ee 


Making the cost at $2.50=$29.70 | And 2.50 is 4 of $20 = $29.70 


Notes. —If the price were $2.75, we would add for 25c. half that at $40c. ora 
tenth of that at $2.50. If the price were $1.75 per ton, we would deduct from 
$23.76 (price at $2) one-eighth of itself, 25c. being 14 of $2; and if $2.25. we 
would add \ of the cost at $2, etc. 

Sometimes 2200 pounds are allowed to aton. In that case, by deducting from 
2200 one-eleventh of itself, or 200, it becomes a ton of 2000 pounds; and by 
making a similar change in the number of pounds to be computed, the fore- 
going rule for net tons can be applied. 


For example, the cost of 1887 lbs. of coal at $2 per ton of 1887 
171.54 
2200 Ibs. would he x off; then moving the decimal point et: 
$1.71546 


three places to the left we get $1.715 + and from this the cost 
at any price. 


230 3 TonNAGE. 


Ner Tons. 


Exam. What is the cost of 1840 lbs. of hay at $13.75 
per ton net ? 


The cost of 1840 lbs. at $20 per ton........ és Ome ta a eee 
and at $10 it is half that at $20...... jis envio ree er ene 
At $2.50 it is one fourth of $10........  b's.6 wie vd oleae 2.30 
and at $1.25 it is half that at $2.50 ...ccocceessees waeaci se Teds 
Making the cost at $13.75 ........,...00- 0.6.4 we oleimite: yop pis Seale tO eee 

Or thus: | 

At $2 per ton the cost is ............000- a ees eee = $1.840 
and at $14 it is seven times that at $2........ oo Sew!) Sa apleme 
At 25c. it is an eighth of that at $2 ........... iste eee .23 
which is deducted, giving the cost at $13.75 ..... eels oy. eee Oe eae 


Nore —If the price per ton were 1334c. = .1375; or $1.387144 = $1.375, the same 


figuring would answer by simply moving the decimal point one place to the left 
in each case. And this method of aliquot parts will be found equally simple at 
any given price per ton by working from the basis given on page 229. 


Exam. What is the cost of 24640 Ibs. of beet sugar at 
$4.262 per ton net ¢ 


Here we have the cost at $20 per ton .... ... .... = $246.40 
and at $4 the cost is one-fifth of that at $20...... eee ae 
At 20c. the cost is $2.46, taken from top figures ..... = 2.46 
and at 63c. it is a third of that at 20c. ............ «karo 82 


giving the cost at 4.205... 2 ses waiene suite eveteeeietem = $92.56 


Nore. —If the price were $4.2724 per ton, we would add for 1c. half of 24c. 
taken at sight from the top figures $246.41; andif $4.2894, the addition would 
be 24c, taken from the top figures, or, rather, 25c, the third figure, 6, being 
greater than 5 or more than one-half, 


a ae ie Ow, iba 


RE ee EF OR eT ae ee ee ee Re 


il aka 


ToNNAGE. 93e 


Gross Tons or Rats to tHE Mite. 


RULE. — To find the number of gross tons of rails to a mile of rail. 
road: Multiply the number of pounds of rail to the yard by 11 and 
divide by 7. 


Exam. How many gross tons of rails to a mile of rail- 
road at 60 pounds to the yard ? 


60x11 __ 660 


Here we have Soe 


= 94.285 gross tons, 


Reason.— The process in full would be: 8 fur. X 40 per X 54 yds. 
multiplied by 60 x 2 (two rails to track), and the result divided by 
2240 (lbs. to the gross ton), or by 40 x 7 X 8, the component factors 


of 2240. Here then we have the fraction 2%10*52x60x?2. 


0x7xe8 > and by 


6x 40x54 x60x2 as 2 x 60. 


cancellation Wax7x5 


Net Tons or Raits to THE Mite. 


Rue. — To find the number of net tons of rails to a mile of rail- 
road: Multiply the number 1.76 by the number of pounds of rail to 
the yard. 


Exam. How many net tons of rails to a mile of road 
at 60 pounds to the yard ? 


Here we have 1.76 * 60 = 105.60 net tons. 


Reason. — The process in full would be 1760 yds. x 60 x 2, di- 


vided by 2000; or WX O*? = THEO S' = 1.76 x 60. 


There are 1760 lineal yards toa statute mile, or, 8 fur, X 40 per. 
x 54 yds. = 1760 yds. 


232 Discounts. 


TRADE Discounts; SHort MeErnops. 


Manufacturers and wholesale dealers usually allow to the trade or 
retail dealers, a reduction from the fixed or list prices of some kinds 
of merchandise. This reduction is called a discount, or a trade dis- 
count. In some lines of business several discounts are allowed, 
Discounts are taken off in succession. Thus, 25%, 15% and 10% off; 


or, as it is generally expressed in business, 25, 15 and 10 off, means 


first a discount of 25%, then 15% of what is left, and finally, 10% of 
the remainder. 


Norts. — The profit on goods is less when 25, 15 and 10 are allowed, than if 50 
per cent were allowed. 

It is immateria] in what order the discounts are taken as it will not affect the 
result. 


Exam. 1. If goods be listed at $65 with 40% off, what 
is the net cost ? 


Usual Method. Short Method. 
$65 X< .40 = $26 $65 x .60 = $39 net. 
then 65 — 26 — $39 


Instead of multiplying the list-price, $65, by .40, and deducting the 


result, it is much shorter to multiply by .60, the difference between 
1.00 and .40, which gives the net at once ; or, as it is said in the trade 
40 off is 60 on. And this short rule will hold good for any series of 
discounts. 


Nors. — Since any per cent. is some number of hundredths, it is properly ex- 
pressed by a decimal fraction ; thus 5 per cent. = 5% = .05. 
It is scarcely necessary to remark that the _list-price represents 100 per cent. 


100 
=100% = ad =1.00=1. Now, if from this we deduct 40% = .40, the difference 


will be 60% = .60 ; so we multiply the list-price by .60 straight to get 40% off. 


3 
x 
q 


Discounts. 935 


TrapvE Discounts; SuHort Meruops. 


Exam. If guods be listed at $3 with 30, 20 and 10 off, 
what is the net cost? 


$3 
4 Or thus: 
7X 8X .9 =.504 
2 1 3 
8 
$1.512 
1.68 
9 
$1.512 


Here we deduct, at sight, 30, 20 and 10, each, from 1002 getting 
70%, 80% and 90%; or, when decimally expressed, .70, .80 and .90; 
or, .7, .8and .9 (the ciphers not affecting the significant figures in 
the decimal expressions), Multiplying by .7, .8and .9 in succession, 
we get $1.512 the net. Or, multiplying .7, .8 and .9 together, we 
get .504 = 50;4% which is the net discount equal to 380, 20 and 10; 
and multiplying this net by the list price we get the net cost. 


Norte 1. To find the net rate of discount equal to several rates: Deduct each 
rate from 100, multiply the differences together and the product is the net rate. 
Thus, 60%, 25% aud 10% off, is equal to 27% net, or to 73% off (.40 x .75 x .90 
= .27); or, (.4 x .73¢ x .9 =.27). 

Nore 2. It should be carefully borne in mind that the cipher, or zero, having 
no value, is used in combinations of figures to fill places where no value is to be 
expressed, and thus to make the other figures occupy those places in which they 
will express the intended values. Hence, a cipher will not affect the value of a 
number unless it be placed between some significant figure and the decimal 
point. Thus, .7, .70, .700, .7000, etc., are all equal in value. But .07, .007, .0007, 
are entirely different, the local value of 7 being changed by the cipher, or ciphers 
coming between that figure and the decimal point. Briefly, then, the use of 
the cipher is to keep the significant figures in proper position with reference to 
the decimal point. 

The use of the decimal point is to mark the place of units; and whether ex- 
pressed or understood, its position is always to the right of units. Itis a matter 
of the utmost importance to have a correct knowledge of making a proper use 
of the decimal point in our calculations. 


934 Discounts. 


TrapE Discounts; SHort Meruops. 


Given the net cost and the discount to find the list-price. 


RuLE. — Divide the net cost by the net discount. 


Exam. If the net cost of goods be $1.512, and the dis- 
counts are 30, 20 and 10 off, what is the list-price ? 


Multiplying .7, .8 and .9 together we get .504, the net discount. 
Then $1.512 + .504; or, moving the decimal point three places to the 
right in these numbers, in other words, multiplying each by 1000, 
to throw off the decimals, we have $1512 + 504 = $3, the list-price. 


Reason. — Since the net cost is found by multiplying the list-price 
by the net discount, we simply reverse the rule here, viz.; divide 
the net cost by the net discount to find the list-price. 


OtrueR Snort Meruops. 


In marking goods, merchants generally take a rate per cent. that 
is an aliquot part of 100, as 50, 334, 25, 124, 84, etc.; and instead of 
multiplying by the net discounts, in such cases, to find the net cost, 
it will be found preferable to make use of the method of aliquot 
parts, as in the following 


Exam. If goods be listed at $240 with 334, 25 and 5 
off; what is the net cost ? 


Here, instead of deducting 334, 25 and 5, each, from 100, $240 


and multiplying by the net discounts, .662, .75 and .95, we 80 
take +, + and =, off in succession (834 = 4 of 100; 26=4 160 
and 5 = 4), thus getting the net cost, $114. 40 
120 
6 
$114 


Notr.— To find the aliquot parts of 100 divide it by 2, 3, 4, 5, ete. Thus, 
100 + 6 = 16%, and 16% is 1-6 of 100, etc. 


Discounts. | 935 


Trape Discounts; SHorrt Meruops. 


Odd rates of discount will be found equally simple in many cases 
where the inexperienced calculator has to make use of long methods. 


Exam. If goods be listed at $148 with 474 off; what is 
the net ? 


Here instead of multiplying by .524, we say 50 is a half, $148 


and 24 is 4, of 50, or 7, of 100; and adding, we have mul- 74 
tiplied by .524. 3.70 
$77.70 


Nore. — And if the rate were 52) off, instead of 4714, the same figuring would 
answer, only instead of adding 3.70 we deduct (100— 471% = 6214); (50 + 264 = 
5236) ; (50—23¢ = 4714). And if the rate were 3714, we would say 25= \& and 124% 
= 1% of 25 (25 + 1256 = 3744); 27% off would be 25= ¥ of 100 and 2¥ a tenth of 
25; and so of other rates. 


| Exam. If goods be listed at $420 with 65% off; what is 
the net ? 


Here we say 65 off equals 35 on and multiplying $420 by $42.0 
.85 we get $147 net. 105. 


$147 


For short method we take 10 and 25; 10% = $42, at sight, and 254 
= + of 420, or 105 which is added in proper position, one place to 
the left. , 


To make the process clearer it must be borne in mind that $420 


SPREE AMIGA LF Gh) Gih 2 otal custo Stal eis min wale ay C6-See Mx once = $420 
eer OE LOS 5. 5 0c's/e s waina a « Aah EEO ie eee = 105 
Ree ite EE LOU. os.d5 a0 Valea vax hese seetc gee ececess = 42 


Making 354 (25 + 10); or (10 + 25 = 85). ...ccceeereee = $147 


236 Discounts. 


TrapE Discounts; SHort Meruops. 


Exam. What is the net cost of goods which are listed 
at $6.40 with a discount of 624% off? 


Here we have 100 — 623 = 374, the net $6.40 
discount. We now say 25=}.......... e ebetence = Pele 
and 124 = 4 of 25, or.40f 100 2... 6.24.04 eee 80 
Adding both results we have multiplied by 374 $2.40 


Or, simply take 4 of $6.40 = 80c. and three times 
80c. gives $2.40, the net cost, 874 being 3 of 100. 


Norsz.—If the rate were 371% off, then the net would be 6214, and to multiply 
by 6244 we would say 50 = 4, and 124=14 of 50. Or, take 14 and multiply by 
5, 6214 being 5 of 100. 


Exam. If goods be listed at $54 with 674 off; what is 
the net cost ? 


Here 100 — 673} = 8234, the net discount $54 X .34 


which is equal to 80 + 24. To multiply by .382} $16.2 
we multiply first by .30, and for 24 we add 1.35 
gy of $54, or 4 of $5.4 ($5.40), 24 being 2, $17.55 


of 100, or + of 10. Or, by looking on .32} as 
.34, simply multiply by .384 bearing in mind that the quarter in 
such cases is in reality 7, of the list price, or } of its tenth. 


Norre.—If the rate were 3214; then the net would be 674, and to multiply by 
.67144 we would multiply first by .60, and to the result add its 14, 714 being & 
of 60. 

Or, take 50, 5 and 1214 (50 + 5 + 12446 = 67.) 

And other odd rates of discount will be found equally simple. If the rate 
were 5714 off, for instance, the net would be .4244. To multiply by .424¢ look 
upon it as .414, bearing in mind that the quarter is 7, asin the foregoing exam- 
ple. If the net were 5734 we would resolve it into 50, 5 and 24; 50=4;5= 5 
of 50, and 244 = 1% of 5. 


pied 
wea. 


Discounts. 237 


TravE Discounts; SHort Merruops. 


Exam. If goods be listed at $80 with 40, 10, 10, 5, 5, 
74 and 3 off; what is the net cost ? 
Since the rates of discount may be taken off in any order without 


affecting the result, we begin here with 7}.and find the net rate as 
follows : 


Deducting 74 from 100 we get 100 
73 
924, and express it decimally, .925 44-Oft;) ==. 925 
2775 
To get 3% off we simply set 3 times oes = eee reo 
4486 
.925 two places to the right and sub- Ga Soe BHO 
4261 
tract. Then 34, of the result is sub- 5 * = 80978 
1D ae ese 
tracted, and 34, of what is left re- 10. Goda 6 
40 ‘© = .3938558x80 
spectively, for the two 5’s, giving $31 .48464 


.80978. Next, 10% is taken off by 
subtracting each left hand figure from the one immediately to the 
right, beginning with the tens, or second figure in .80978, thus: 
7 from 8 leaves 1; 9 from 17 leaves 8; carry 1 to 0; 1 from 9 leaves 
8; 8 from 10 leaves 2; 1 tocarry taken from 8 leaves 7, the result is 
.72881. And from this number the second 10% is got in like man- 
ner, giving 65593. Then 40% is taken off by multiplying .65593 by 
.6, the net equivalent to .40; the product, .393558 is the net rate. 
This is now multiplied by 80, the list price, and we get $31.48 +, 
the net cost. 

Nots.—The net rate, .393558 is nearly equal to .39;°,5,, or 393%, and is the 


product of .9214 x .97 x .95 x .95 x .90 x .90 x .60, the multiplication being per- 
formed much more simply by the process given above. 

Ry taking 3954,% from 100 we get 603%% which is the rate of discount off 
equivalent to the seven rates, 40, two 10's, two 5’s, 744 and 3 off. (See page 238, 
also note 1, same page.) 


INTEREST REVIEWED. 


Since the interest of $100 for1 yearat1% = a ae | 
the interest of $100 for 100 years at. 1% $100. 
in other words, money doubles, or the interest ) 
will equal the principal in 100 years, at 1%, simple interest. 

If, then, 100 years be divided by the rate per cent. it will give the 
time, in years, when money doubles at that rate, simple interest. 

Thus, any sum of money at 2%, will double itself in 50 years; at 
4%, in 25 years; 5%, in 20 years; 6%, 162 years, &c. 

Hence, taking 100 years and 14 for the basis, and reducing the 

100 years to months and days, we have the time in which 
money doubles at 1% = 36000 days = 1200 mo. = 100 years. 


“A fe 6° 2:9 -= 18000:°.*. = - 600 eee 
ws es “© 247 = 16000 ‘“*. = 60355 ee 
ee cs ‘63% = 12000. -** 2 540023 ree 
& ff “ 38¢ = 9600 ‘“ = 820) 5 ee 
a a “44% $9000 -“ = 3000 ee 
&¢ As ‘| 44¢ = 8000“: = (2662 ae 
* a ‘5G = .7200- 8. == 22a e 
fs £5 666.9 = 6000: . >= 200) i See 
ae aS 98% ==. 4800 =. £ se 2 AGN ee 
Be af "8 +=. 4000. 2% == 100 eee 
‘s ef ‘94 = 4000.) “ + =).130) See 
= . © 10:'9:=.. 8600) “te 120 Sea 
ue “f "12-4 = -3000.. 4%" = 100 te ee 
&e. &e. &c. &e. 


Nore.—-It will be seen, on examining the foregoing, that the time is pro- 
portional to the rate, and vice versa; thus, the interest of any sum of money 
for 18000 days at 2%, is equal to the interest of that snm for 9000 days at 47; 
the interest at 9% for 4000 days, is equal to that at 44%% for 8000 days, or 2144, 
for 16000 days; and the interest of any sum for 4800 days at 714%, is equal to 
the interest for 9600 days at 3%%, &c 


Interest REVIEWED. 939 


A careful analysis of the matter given on the foregoing page, will 
now enable us to compute interest at any rate, and for any time, 
with ease and rapidity. Todoso, we must keep in view the basis, 
or the time in which money doubles at 1%, simple interest, viz., 
36000 days, 1200 mo. or 100 years. 


Take any particular rate, say 24%, for example, and suppose it 
were required to find the interest of $5764.50 for 16 days, at that 
rate. . 

Here we simply point off three $5 7645 
figures to the right, counting from 
the decimal point, and we get $5.7645, the required interest, true to 
four places of decimals. 


Reason. —Since money doubles in 36000 da. at 1%, simple interest, it will 
double at 214% in 16000 da.; 214 being contained 16000 times in 36000. Hence, 


The interest of $5764.50 for 16000 da. at 214% = $5764.50 
one tenth of this is the interest for 1600 ‘‘ os = $576.45 
one tenth of the latter is the int. for L6G 2s ne = $57 645 
and a tenth of this last is the int. for ‘Gass ey = $5. 7645 


We have here, now, a basis to find the interest for any number of 
‘days, at Q1¢, If 40 days, take + of 160 days’ interest, or of $57.646; 
the interest is $14.41; if 80 days, the interest will be half of 
$57.645, or $28.82 ; if 32 days, it will be two times 16 days’ int. or 
$11.529; if 10 days’ interest be required, take 4 of 160 days’ interest, 
then 4 of that half; or, take 1 of 160, then 1 of that fourth ; the 
result will be 10 days’ interest. 

If the time be 8 mo. 10 da. or 100 days, we have 6 times 16 da. 
equal 96, plus 4 da., or} of 16; and if 4 mo. or 120 da., we have 
160 da. less a quarter of that, or 40 days’ interest, the difference will 
be 4 mo., or 120 days’ interest, &c. 

And if one day’s interest be required at 24%: point off three 
figures from the right, counting from the decimal point, the result 
is 16 days’ interest always. Then take 4 and } of that $ (2x8=16.) 
or, take } and } of that } (4x4=16.) 


940 IntTEREST REVIEWED. 


Again, suppose it were required to find the interest on a loan, say, 
of $300,000 for 1 day, at 334. 
By referring to page 238, we see that 3? is contained 9600 times in 
36000 days. 
~ Now, since the interest will equal the principal in 9600 days, at 
33%, simple interest, always : 


We have, here, at sight, the interest for 9600 da. = $300000 
one tenth of this, or the interest for 960 “* = .$30000 
one tenth of the latter, or the interest for 96 ‘* = $3000 
and x, of this last is the interest for a $31.25 

Hence, 


RuLE.— To find the interest of any sum for 1 day, at 38%: Move 
the decimal point two places to the left, in other words, take 1% of the 
principal, and the result is 96 days’ interest, always. Divide this by 
96 for 1 day’s interest. 


Exam. What is the interest of $376860.48 for 1 day 
at 33% ? 


Taking 1% of the principal we have $3768 .6045 = 96 da. int. 
one twelfth of $3768.6048 gives $314.0504 = 8‘ * 
and one eighth of $314.0504 gives $29. 2563 =" 1" 


Notrs.—1. It will be observed that, in dividing by 96, to get 1 day’s interest, 
we have made use of 12 and 8, the component factors of that number — 
(12 x 8= 96.) And in all such cases where the divisor can be readily factored 
this method of division is to be preferred. Thus, to get 1 day’s interest at 714¢, 
we would divide 1% of the principal by 48 (the significant figures in 4800 days, 
or the time in which money doubles at 7147, simple interest), making use of the 
factors, 6 and 8; and if 5%, we would divide 1% of the principal by 72 (the sig- 
nificant figures in 7200 days, the time in which money doubles at 5%, simple in- 
terest), using § and 9, or 6 and 12, the factors of 72, &c. 

2. If we examine carefully the foregoing example, it will be readily seen how 
easily the interest for any number of days may be obtained at 334%. If the time 
were 48 days, for instance, instead of multiplying 1 day’s interest by 48, we 
would simply take 44 of 96 days’ interest, or 6 times 8 days’ interest. 32 days’ 
interest is 4 of 96 days, or 4times 8 days’ interest; 33 days would be 82 plus 1 
day, and 81 days would be 32 minus 1 day. By adding 1 day and 8 days’ interest 
we have 9 days, and by deducting 1 from 8 we have 7 days, &c. 


Interest REVIEWED. 241 


It will now be seen that, when the rate is an exact divisor of 
36000 days, or 1200 mo., the computation of interest can be made 
simple and interesting. And this being properly understood, com- 
putations will be foun‘ equally simple when the rate is not an exact 
divisor. Take for instance, the following 


Exam. What is the interest of $75684 for 1 day, at 
23% ? 


Since 234 is not.an exact divisor of 36000, we take 8% which is contained 12000 
times in 36000 ; and since the interest of any sum for 12000 days at 3%, is equal 
to the principal, the interest for 12 days is =,)55 of the principal; and this is 
found by simply moving the decimal point three places to the left in the given 
principal always. 


Here, then, we have 12 days’ interest.......... eo $75 . 684. 
and +, of this is the interest for 1 day at 3%........ = %6. 307 
Now, 37 is 7 of 3%, and we deduct 7y............. = 525 
making the interest for 1 day at 23¢..... ..... ee — $5. 782 


Note.—If the rate were 344%, we would, of course, add as instead of subtract- 
ing ; and if 314% then } of 3% would be added, &% being 4 of 3, &c. Hence, 


Rue. Zo find the interest of any sum for 1 day, at any rate per 
cent. : Divide 86000 cays by the rate, if it be an exact divisor, divide 
the yiven principal by the result, and the quotient is 1 day’s interest. 
Tf the rate be not an exact divisor, take the nearest rate which is an 
exact divisor, find the interest at that rate, and add, or subtract, for 
the difference. 

If the rate be 47, for example, 4 is contained 9000 times in 36000 and we divide 
the principal by 9000 to get 1 day’s interest. If 414%, the divisor is 8000; if 67, 


the divisor is 6000; and if 5%, it is 7200, &c. 

Tn dividing, point off as many figures from the right, in the given principal, 
as there are ciphers in the divisor, cownting from the decimal point alivays. - 
Thus, to divide by 9000, point off three figures and divide by 9; for 6000, point off 
three and divide by 6. In dividing by 4500 for 8%, point off two figures and 
divide by 45, using the factors 5 and 9; first taking + of the principal, ther tof 
that fifth, making nse of short division in all such cases as pointed out in the 
example given on page 240. 


249 INTEREST REVIEWED. 


The 6% method is used in most of the Banking Institutions of the 
country (when interest tables are not used), and is taught in the 
majority of Commercial Colleges, as the shortest method of comput- 
ing interest, a very excellent method, indeed (given on p. 180 of this 
work.) 

In adhering to this method, however, a good deal of unnecessary . 
labor has to be gone through in many cases where a knowledge of 
the foregoing methods will frequently give the interest, at sight, 
without figuring, at all. 

Take, for instance, the following problem at 447, and compute the 
interest on the basis of 64. 


Exam. What is the interest of $8000 for 149 days at 
41%? 


Nots.—As a general rule 60 days’ interest is taken as the basis of calcula- — 


tion, and the solution, by the 6% method, would be something like the 
following: - 


Pointing off two places, we have.... $80.00 = G0 da. int. 
multiplying this by 2, we have........ . $160,000 3426s 
one third of 60 days’ interest, or...... 26.66°es 1 200 ieee 
one tenth of 60 days’ interest, or...... 8.00 == 6 en 
and one half of 6 days’ interest, or.... 4.00) = 8 eee 
giving the interest for 149 da. at 6% = $198.66 = 149“ « 
From this we deduct } to get.......... 49 .66 
44%; 14% being 1 of 67................ $149.00 = int. at 444. 


Note.—Too many figures even for the 6% method. Better point off one place, 
and we have $800.0 = 600 days’ int. at 6%; 150 da. = 4 of 600; less 1day, or tof | 
$8; the difference is 149 days’ int. at 6%. Then 1 off is 442. 


_ But either method is too long in this case if it be borne in mind — 
that the interest of $8000 for 149 days, is the same as the interest of 
$149 for 8000 days, and that 8000 days is the basis ; in other words, | 


INTEREST REVIEWED, 243 


that the interest will equal the principal in 8000 days, at 447, simple 
interest, always. 

Hence, 

The problem can be reversed so 2s to read: What is the interest 
of $149 for 8000 days, at 45%? 

The answer is at sight, namely, $149. 


Notr.—This method of reversing the problem can always be used when it is 
more convenient to take the dollars for the days. 


If, for instance, it were reguired to find the interest of $1000 for 
149 days, at 44%, we would reason thus: 


The interest of $1000 for 149 days being = $149 for 1000 SG. , 
and since the interest of $149 for 8000 da. =35 . $149 
4 of this will be the interest for 1000 ‘* = $18.625 


And, in like manner, the interest of $2000, $3000, $4000, $5000, 
$6000, $7000, $9000, $10000, $12000; or any multiple, as $20000, 
$25000, $48000, &c., or any part, as $4, $40, $400, &c., can be readily 
obtained. 

To make this important reethoe more clear let us take another 


Exam. What is the interest of $4500 for 5 mo. 17 da. 
at 8%? 


In 5 mo. 17 da. there are 167 days. 

Now, since money doubles in 4500 days, at 8%, simple interest, 
and that the interest of $4500 for 167 da. = that of $167 for 4500 
da. the required interest is at sight, viz., $167. 


Notre.—If the principal were $2250, the interest would be % of $167 = $83.50; 
if $9000, the interest would be twice $167, or $334; if $1500, the interest would 
be &% of $167, or $55.67 ; and if $15000, the interest would be 10 times that, &c. 

And if the rate were 7%, we would, in this case, deduct } of the int. at 8%. 

Since 7 is not an exact divisor of 36000; interest can be computed at 6% and 4 
added, or at 8% and } deducted, whichever is most convenient. 


944 InTEREST REVIEWED. 


Should the time be given in months; years and months; or years, 
months and days, it will be found preferable, in many cases, to take 
the time in which money doubles, in months, as the basis of calcu- 
lation, instead of days. Thus: 


Exam. What is the interest on a bond of $5000 for 
{ mo, Oda. jal oy? 


Notr.—lInstead of taking 7200 days, the time in which money doubles at 5%, 
for the basis of calculation, we prefer to take 240 mo , its equivalent ; and since 
the interest of $5000 for 240 mo. at 5%, is $5000, the interest for 24 mo. is +5 of 
that, or $500. 


Here, then, by pointing off one place $500.0 = 24 mo. 


in $5000, we have 24 mo. interest, or $500 $126; + == Gee 
6 mo. =} of 24 mo., + of $500 = $125 $20.83 =="2 bes 
1mo. =tof 6mo.; tof $125 = $20.83; $6.25 = Qda. 
and for 9 days, we take 4 of $50, or 72 da. $152.08 


interest, which is always at sight by point- 

ing off two places in the principal, $5000. Or, to get 9 da. interest 
we could take 6 da. =+4o0f 1 mo. and 3 da.=4of 6. Or, better 
thus, perhaps : 


Reducing the time to days, we have 7 mo. 9 da. = 219 da. and 
since the interest of $5000 for 219 days, is the same as the interest 
of $219 for 5000 days, the problem can be reversed so as to read: 
The interest of $219 for 5000 da., at 5Z, and the solution be obtained 
as follows: 


The interest of $219 for 7200 da., at 5%, being.... = $219 
10 times this will: be:for “72000. ; <<’ -.vae.4 eee ee = $2190 
Now, since the interest for 72000 da. = $2190 
zs of this interest will be that for 6000 << = $182.50 
4 of this will be the interest for 1000 ‘“ = $30. 42 


and the difference will be for 5000 « = $152.08 — 


sf See eae 


InteEREST Rui VIEWED. Y45 


Rue. When the rate of interest changes frequently : Find the rate 
for 1 day, and compute the interest on the given principal at that rate. 


Exam. What is the interest on a loan of $35000 made 
on Jan. 1, and paid off on the 15th, the rate of interest 
changing as follows: 


Jan. 1 24% 
3 2% 
6 3% 
Pir ake 
9 3% 
10 35% 
12 2% 
15 paid off. 


Here $35000 bears interest from the Ist. to the 3rd, or two days at 
24%; next, 3 days at 2%, &c. Now, 237 for 2 days = 54 for 1 day, Xe. 
Hence the process : 


24 X 2=5 
Seo a 
Bee == 8 
34 xX 2=7 
3 ea 
ye er 
3, <2 16 


07% for 1 day. 


We now compute the interest on $35000 for 1 day at 872, or, what 
amounts to the same thing, the interest for 37 days at 12. 
And since the interest of $35000 for 36000 days at 1% == $35000 


sé 6é 66 66 ce 36 days “é Lad a $35 
6é ce ee 6é 6< 1 day cé¢ 66 = 97 
giving the interest required ‘* 837 days a ey $35.97 


Norz.—In getting J] day’s int. take } of $35; then % of the result, making use 
of the factors, 6 and 6 (6 x 6 = 36.) 


ANNUAL INTEREST. 


Rue. When interest is payable annually : 
_ I. Compute simple interest of the principal from tuts given date to 
the date of settlement. 

II, Add to this the interest of each year’ s interest from the time of 
its accruing to the date of settlement. 


Exam. What is the amount due, in 3 years, 6 months, © 
on a note of $8000, interest payable annually, at 5%? 


$8000 x .05 = $400 = Interest for 1 year. 
$400 x 34 = $1400 *s ‘« 35 years. 
$400 x .05 x 2h = $50 ‘* on the 1st year’s interest 


te | 


$400 x .05 X 1$= 30 “ $6 OSes 
$400 x .05x 4= 10 i «6c. Oras 
8000 Principal. 
$9490 

In this the interest of $8000 for 3% years, at 5% 
Next, the interest of $400 ‘“* 24 - 
then ¢< 66 66 ee (a3 14 6s ce 
finally ce 66 “6 anf 66 q year 66 

Making the entire interest................. 


414 = $90. 


Tt 


II 


“6 


oe 


$1400 
50 
30 
10 


$1490 


Notr.—It will be seen that the interest of $400 for 244 years, 14 yrs. and& 
yr., is equal to the interest of $400 for 44% yrs., their sum. Thus: $400 x .05 x 


Pei | Aly FeASY MCR N'TS, 


A Partial Payment is payment in part of a note, mortgage, bond, 
or other obligation. 

An Indorsement is an acknowledgment of payment, written on the 
back of the note, mortgage, &c., stating the time and amount of the 
payment made on the obligation, 

NotrE.—The United States Rule, given on page 129 of this work, has been 
adopted by nearly all the states of the Union, to secure uniformity in the 
method of computing interest where partial payments have been made on 
bonds, mortgages and other obligations. It may, however, be proper, here, to 
give the reader a knowledge of the method generally used in business for sete 
tling notes and interest accounts, and is called 


Tne MERCHANTS’ RuLE.—I. Compute the interest of the principal 
from its date to the time of settlement, and add wt to the principal. 
Il. Compute the tnterest of cach payment from its respective date to 
the time of settlement and add this interest to the payments. 

Ul. Krom the amount of the principal take the amount of the pay- 
ments, and the remainder will be the,balance due. 


EXAMPLE. 
$5000 Albany, N. Y., Aug. 1, 1902. 


Sixty days after date I promise to pay to A. B., or order, five 
thousand dollars, with interest, value received. 
BOOZ. 
Indorsed as follows: Nov. 1, 1902, $500; Feb. 1, 1903, $500; 
April 1, 1908, $1000; July 1, 1903, $2000. How much was due 
Aug. 1, 1903? 


Process: 
Principal = $5000. Payment = $500 

lyr. int. at6% = 300 9 mo. int. at 6% = 22.50 

$5300 Payment =a O00: 

Deduct 4057.50 |, 6 mo. he at 6% = ae, 

Wiener. aymen = P 

Balance due = $1282.50 4 mo, int. at 6% = 90). 

Payment = 2000 

1 mo. int. at 67 = 10 


$4067.50 


948 Hints anp Hewp ror tHe STUDENT. 


. . .865)1709749265(4684244.5616 
...730| 2497 
3074 


In long operations in division, it 
or where the same divisor is to 2 
be frequently used, zt will be 8. .19095 1549 
found advantageous to form a 4..1460} »§ gg2 
table of the several products of 5. .1825 1626 
the divisor and the nine digits. 6. .2190 1665 
Thus, to divide 1709749265 by 7. .2555 sete: 
365, say to four places of deci- 8. .2920 600 
mals, we look in the table and 9. .38285 2350 
find that 1460, the product by 
4, is the nearest below 1709; we place 4 in the quotient, and, 
without setting down 1460, subtract it from 1709, setting down the 
remainder 249, to which 7 is brought down making 2497. We then 
see in the table that 2190, the product by 6, is the nearest below 
2497; 6 is put in the quotient, and 2190, the product by 6, is the 
nearest below 2497; 6 is put in the quotient, and 2190 is subtracted 
from 2497 and the remainder is 307, to which 4 is brought down, etc. 
To divide by 365, however, we prefer to proceed as follows: 
Annexing a cipher, and doubling the re- 17097492650 
sult, we have 84194985300. ‘T'o this we add 34194985300 
its third, one tenth of that third and one 11398328433 


tenth of that tenth: the sum is 46847129860, 1129832843 
which we divide by 10001, according to rule 113983294 
II, page 15. 


- To divide by 10001, we simply cut off  4684712/9860+10001 
four figures to the right for 10000; then 
deduct once 4684712, the figures on the left 
of the line, and to the result add 468, found 
on the left of the line, also; the quotient is 
4684244 .5616, true to four places of decimals. 
(See rule II, page 15.) 

Reason.— By annexing a cipher to 865 and doubling 3650 
the result; then adding to this its third, one tenth of 7300 
that third and one tenth of that tenth, we obtain 10001 for - 24334 


.0616 


a simple divisor, thus establishing a simple method for 2431 
dividing by 3865, always. (See examples pp. 51, 52 244 
and 58.) aes 


Hints AND HELPs FOR THE STUDENT. 249 


By simple divisors we mean 10, 100, 1000, 10000, etc., or any 
number which isa little more, or a little less, as 103, 1008, 1040, etc., 
89, 989, 91, 94, 995, 9996, etc., also, the multiples, submultiples and 


aliquot parts of these numbers. 

Divisors may be simplified by any process that will make them 19, 
100, 1000, 10000, etc. Thus, to divide by 74, 7.5, 750, 7500, etc.: add 
to each its third, and the numbers become 10, 10, 100, 1000, 10000, etc. 

Exam. 1. In 2592 pounds of oil, how many gallons, 
74 pounds to the gallon ? 


Here, instead of dividing 2592 by 74, we add to each its 2592 
third, and we have 3456+10, which gives 345.6, the re- 864 
quired number of gallons. 345.6 


Exam. 2. If a gross of articles cost $237.60, what is 
the cost of a single article? 


Instead of dividing by 144 (articles in a gross) $1/663.20+-1008 
we multiply it by 7, getting 1008 for a simple 8 


divisor, then, multiplying $237.60, also, by 7, 651520 
we have 1663.20+1008; and the answer is $1.65. ; 520 
(See the rule and exam. page 57.) ede 


Exam. 3. Divide 35886432 by 9524. 
By setting half of each number, in this 35886432 9524 


example, one place to the right, and adding, © 17943216 4762 
3768. 5236 

And if 4762 were the divisor; 2 times that ee 
number set one place to the left, and added, 
would make 10002. 

Exam. 4. Divide 3229184 by 476. 

Setting 2 times each term one place to the 3229184 476 
left, and adding, we get 67812864+-9996; the 6458368 952 


quotient is 6784. (See exam. 8, page 23.) 

And if the divisor were 952; half that 
number set one place to the right, and added, 
would make 9996. 


6781 |2864-- 9996 
2\7124 
8 


6783 9996 


250 Hints AnD HeEtp For THE STUDENT. 


Let it be carefully borne in mind that, when a divisor can be 
simplified, any multiple, submultiple, or aliquot part of it, can also 
be simplified. 

Thus, 167 multiplied by 6 gives 1002 for a simple divisor. Now, 
if we take the multiples, 334, 501, 668, 835, 8350, 1169, 1836, 18360, 
1508, 15080, etc., we have: 384x*8=—1002; 501*2=1002; and by 
adding to 668 and 835, one half and one fifth of each, respectively, 
we get 1002. ‘To simplify 1169, 1886 and 1508, we deduct from 
each, one seventh, one fourth and one third respectively, to get 1002. 

And if we take the submultiples, 834, 552, 413, 332, 278, 288, 20%, 
183, etc., we multiply by 2, 3, 4, 5, ete., to get 167, which, in turn, 
is multiplied by 6, to get 1002, the same simple divisor for all the 
numbers. (See page 70.) 

Again, many numbers, which, at first glance, may appear difficult 
to simplify, will be found, on slight inspection, to be easily managed, 
and frequently by a choice of several methods. ‘Take for instance, 
69, 690, 6900, etc., and their multiples, 1388, 1380, etc., 276, 2760, 
27600, etc. 


Exam. Divide 50950980 by 690.. We give the solution 
of this in four different ways, as follows: 


First: In this, we add tothe terms one half 50950980 690 
25475490 345 


76426470 1035 


of each, respectively; then, setting said half 


one place to the right under each result and 25475490 345 
subtracting, we have 733789210+10005; and oe est 
the quotient is 73842; the remainder being 7384119820 
equal to 1 ; 185 

10005 


50950980 690 
Second: Here, we add to the terms a third of 16983660 230 


each, respectively, and we have 67934640 to be pti ees 

divided by 920. Cutting off the cipher from each, 434 72 

we have 6793464+-92; and the quotient is 78842. - 

(See exam. 6, 7 and 8, p. 23.) ae 
73841'92 


Nots. —It is well to remember that, whether we use the method of. addition 
or that of subtraction, the figures on the left of the line, only, are to be operated 
upon; andif any figure be carried from right to left, over the line, in either 
case, such figure must also be operated upon. In this last example we multiply 
5 (2 plus 3 carried) by 8, to get 40; and in the other example, 37 (86+1) is multi- 
plied by 5 to get 185. 


ae Pe Sy eae 


Hints AND HeELrs ror THE STUDENT. 251 


es 

Third: Here, we take 700 for approximate 509509180 + 690 
divisor; the difference is 10. Cutting off two 
figures and dividing by 7, divides by 700. 


T2787|118 10 


We next add +o, or yo of each partial 1039 oe 
quotient, till all these quotients are exhausted 14 843 
(divide by 7 and set each result one place to se 

7 


the right) using the fractional form for the eae aati 
remainders. The sum of the several results 73841|984 = 73842 
is 73841 | 894, which is equal to 73842, the 

required quotient. : 

Dividing 50950980 by 690, it is scarcely necessary to say, is the 
same as dividing one seventh of the one by one seventh of the other, 
or dividing 72787112 by 984 (690+7=984) and, therefore, the re- 
mainder 987 being equal to the divisor, 1 is added making 73842. 

Again, in the addition, 1 is carried from the figures on the right, 
to those on the left of the line, and if written down, the carried 
figure would be set under 14: the carried figure, therefore, (1 under- 
stood) forms a partial quotient the same as the numbers immediately 
above it; and , of this figure, .013, must be added. 

Now, since 98; is the simplified divisor, 13 (100—984) is the com- 
plement, or multiplier: but multiplying by 13 is multiplying by 19; 
and since we have already divided by 100, by means of the line, and 
now dividing by 7, we have added the +, or +5, in each case. 


509509/80 


; sath Pega 72787|11.428571 
Fourth: The process, in this, is the same as 1039'81.428571 


in the last example, except that the remainders 14/84, 285714 


are treated decimally instead of fractionally. s 428571 


73841|98 .571428 


The remainders, in the present instance, run into circulating 
decimals, or periodicals, and if carried out, their sum would be 
-571428, repeated. Now, if 4, the fractional part of 984, be reduced 
to a decimal it will be found to be equal to .571428, also, repeated. 
The divisor 984, therefore, is equal to 98.571428; and since the 
remainder is equal to the divisor, 1 is added to the quotient, making 
73842, as in the other cases. 


952 Hints AND HELPS FOR THE STUDENT. 


From the examples and illustrations given, it will now be readily 
seen how to divide by 29, 39, 59, 79, 390, 399, 599, 59990, etc. 

Again, if we take such numbers as 200, 2000, 800, 30000, 400000, 
700, 7000, etc., and deduct the significant figure from these numbers, 
and also from the successive remainders, each remainder thus found, 
can be simplified when used for a divisor. 


Take for instance, 7000, and deduct 7 from it, and also from each 


successive remainder, and we find that 69938, 6986, 6979, 6972, 6965, 
6958, etc., can be simplified. 


If, for eae 6986 were presented for aimee we would com- 


pare it with 7000, as shown in the margin. 

Here we see that 14 is the difference between 6986 and {6 
7000, and that 7, the significant figure of 7000, is contained 6986 
in 14, an exact number of times; and this being understood, —_—— 
7 will be contained in 6986, also, an exact number of times: 998 
and so we divide 6986 by 7 to get 998 for a simple divisor. 


Exam. Divide 76874626 by 5994, exact, to six places 


of decimals. 


‘In this, we see at a glance that 6 76874626 6... 
is the difference between 5994 and : 37.6’ +999 5994 
6000; hence 6, the significant figure ” ue 999 


in 6000, is contained in 5994 giving —_jag55/ag9 GEaiaGG 


999 for a simple divisor. We then 262/666 
divide the dividend by 6, getting 263 
12812437.6’ which is divided by 999. . 262929|595 


In dividing by 6, we treat the remainder decimally, and obtain 
the repetend .6’, in other words, 6 is repeated to infinity, and in get- 
ting the decimal part of the quotient, 6 is repeated six times for the 
six decimal places required. The quotient is 12825.262929, true to 
six places of decimals. 


Nore. — And if the divisor were 6 less than 5994, that is, 5988, we would divide 
5988 by 6, getting 998 for a simple divisor; and so with the other numbers differ- 
ing by 6. Thus, 5982-+-6=997. ete. 


Hints Ano HeEtps For THE STUDENT. 253 


When one part of the divisor is a multiple of the other part, which, 
in itself is a simple divisor, the division can be simplified. 

Take, for instance, such numbers as, 96192, 960192, 960120, etc., 
93372, 930372, 48144, 480144, 36108, 360108, 24096, 240960, etc. 

Suppose, now, it were required to divide by any one of these, say 
930372. Arranging the numbers as follows, the component factors 
will be readily seen: 


96 96 93 93 
96192 960192 93372 930372 
1002 10002 1004 10004 


Exam. Divide 30207318096 by 930872. 
93 


A moment’s glance, here, shows that 3020731] 8096930372 
1208} 2924 10004 


372, one part of the divisor, is 4 times 93, eee 
3019523} 5172 


the other part, in other words, 93 and 4832 
10004 are component factors of 930372 10004 
(93>< 10004=930372). We now divide, first, a be nga 
by 10004, getting 3019524 for quotient, 147/91 
which in turn, we divide by 93, and the 10 sh 


quotient is 32467 | 93=32468. 39467 93 

Again, if these numbers be reversed such as, 19296, 192960, etc., 
the division can also be simplified. To divide by 192960, 
for instance, we would take half, or 96480, and make use of 96 
the component factors 96 and 1005, as shown in the margin, 96480 
dividing, first, by 1005, and the result found, by 96: or, 1005 
divide, first, by 96 and next by 1005. 

And if 48144 were presented for divisor, we would divide 48 
by 48, using the factors 6 and 8 (6X8=48); then divide the 48144 


result by 1003: and so with the others, and similar numbers. 1003 
Or, doubling 48144 gives 96288, the component factors 96 
being 96 and 1003. (See exam. and note page 79.) 96288 

1008 


Notr. — This method of using the component factors, in connection with the 
simplified methods given in this work, will be of great advantage when the 
divisor is not subject to change. (See page 55, exam. and note.) 


254 Hints AND HELps FOR THE STUDENT. 


DECIMAL FRACTIONS, OR DECIMALS. 


A Decimal Fraction, or a Decimal, is a fraction having 10, or some 
power of 10, such as 100, 1000, 10000, etc., for its denominator. 


3 9 4756 j 
Thus, #, zs. azodsoo 4436, are decimals. 


In the notation of decimals, the denominator is usually omitted, 
the value of the fraction being expressed by pointing off as many 
decimal places in the numerator as there are ciphers in the 
denominator. 


Should there not be a sufficient number of figures in the numer- 
ator, the deficiency is to be supplied by prefixing ciphers. 


Thus, 335, +30, rovooo 4338, expressed in the notation of decimals, 


are written, respectively, .8, .09, .00003, 4.756. Hence, conversely: 


The denominator of a decimal thus expressed, is unity, or 1, 
followed by as many ciphers as there are figures in the decimal. 


Thus, .57 is 75%, .004 is e475, and .00063 is +yS2q5- 


It will be readily seen from this notation, that the figure im- 
mediately to the right of the decimal point, is tenths, the next, 
hundredths, the third thousandths, etc. 


Thus, since 376 is equivalent to Bee a the fraction .376, or 


pins is equivalent to #oy>5-+2rh00 +1000 OF otra tri: 


The values of figures in decimals, as in whole numbers; are in- 
creased in a tenfold degree by removing the decimal point one place 
towards the right hand, and are diminished in a like degree by 
removing the point one place to the left. 


Thus, in the decimal .003, by removing the point one place to the 
right we have .03, which denotes ;3,, or 7395, and is therefore ten 
times the given fraction, .0038, or +29; but by removing the point 
one place to the left we have .0003, or +5255, which is only a tenth 
part of .003, or zp, OF T0000: 

Hence, 

A decimal is multiplied by 10, if the point be removed one place 


Hints and HeEtps ror THE STUDENT. 255 


to the right; by 100, if two places; by 1000, if three places, etc., 
and, conversely, a decimal is divided by 10, if the point be removed 
one place towards the left hand, by 100, if two places; by 1000 if 
three places, etc., vacant places, when there are such, being supplied 
in both cases by ciphers. 


Thus, .6845x10=—6.345, or 6,343,;; 6.345 x 100=—634.5; 6.31000 
— 6300. Also, 78.48+10=7.848; .784+100=.00784; 7.3+100= 
.073, etc. Hence, 


The value of a decimal is not changed by annexing a cipher to the 
end of it, nor by taking one away, a cipher not affecting the value 
of a number except when placed between a significant figure and 
the decimal point. (See note 2, page 233.) 


Thus, .50=.5=.500—.5000; each being equivalent to one half. 


But if a cipher be placed between the significant figure and the 
decimal point, the value of the number is changed at once. 


Thus, 50., 500., 5000., .05, .005, .0005, ete. 


From the foregoing view of the nature of decimals, it will be seen 
that there is, in every respect, the closest resemblance between 
them and whole numbers; and hence all operations on decimals are 
performed exactly in the same manner as those on whole numbers, 
due attention being paid to the position of the decimal point. This 
last circumstance, indeed, requires the utmost care when making 
our calculations, as the point is the characteristic of the decimal; 
and, from what precedes, it is evident how much depends on its | 
proper position. 


Rute I. To reduce a common fraction to a decimal, in other words, 
to divide a smaller number by a larger: First, annex a cipher to the 
smaller number and divide by the larger, and to the significant figure 
or cipher found in the quotient, prefix a point. Then, if there bea 
remainder, annex ciphers, and continue the division till nothing 
remains, or till the result consists of as many figures as may be 
deemed necessary. 


256 Hints ann Hewps ror THE STUDENT. 


Exam. Reduce 515, to a decimal, in other words, 
divide 15 by 2560, expressing the quotient decimally. 


Here, by annexing a cipher 15 becomes 2560)15000(. 005859375 
150, in which 2560 is not contained; and 12800 
therefore a cipher is placed in the quotient. 
Annexing another cipher, 150 becomes 1500, 
in which 2560 is not contained, and another cipher is put in the 
quotient. After this, the division proceeds in the usual way, a 
cipher being added each time; and the quotient, or decimal is found 
to be .005859375, or 728) %3eo0 : Which by reduction to its lowest 
terms would become »}3,, the given fraction, thereby proving the 
work to be correct. (The work is left for the student to perform.) 


etc., etc. 


Nots.—It may be well to remark that, when the fractional form of the decimal 
is used, the point is always omitted, the numerator of the fraction consisting of 
the significant figures of the decimal, while the denominator will always be 1, 
followed by as many ciphers as there are places in the decimal, including signifi- 
cant figures, and ciphers, if there be any. 


RuLeE Il. To divide decimals: If the number of decimal places in 
the divisor and dividend be not equal ; make them equal by annexing 
ciphers to the one having the least number. Then reject the points and 
divide as in whole numbers, and if the divisor be contained tin the 
dividend, the figure, or figures found in the quotient will be a whole 
number. Having used the last figure of the dividend, annex ciphers, 
if there be a remainder, and continue the division till nothing remains, 
or till the number of figures considered necessary is found in the 
quotient. The part of the quotient thus obtained, will be a decimal, 

If, after rejecting the points, the divisor be greater than the dividend. 
the work will proceed according to the rule given on page 255, 


Exam. 1. Divide 2738.5 by 78.54. 


Here, by annexing a cipher to the dividend, we make the number 
of decimal places in both numbers equal. Then rejecting the points, 
we have 273850 to be divided by 7854; and we find the integral part 
to be 34. Annexing ciphers, and continuing the division we get the 
decimal part .86758, etc. The quotient, therefore, is 34.86758, etc. 

(For simple method of dividing by 7854, see p. 56.) 


Hints and Henps FoR THE STUDENT. 257 


Exam. 2. Divide .1342 by 67.1. 


In this, we equalize the number of decimal places in both numbers, 
by annexing three ciphers to the divisor; and rejecting the points, 
we have 1342 to be divided by 671000. Then, the divisor being 
greater than the dividend, we proceed according to rule I, page 208, 

The required quotient is .002. 


MULTIPLICATION OF DECIMALS. 


Rue. — To multiply decimals: Multiply the factors as in simple 
multiplication and point off in the product as many places of decimals, 
as there are in both factors ; supplying the deficiency, when there is 
one, by prefixing ciphers. 


Exam. 1. Multiply 66.3 by .582. 


Here, we multiply 663 by 582 as whole numbers; the product is 
385866. Then, because there are three decimal places in one factor, 
and one in the other, there must be four decimal places in the 
product 38.5866. 


Exam. 2. Multiply .14 by .6. 


Here, a cipher must be prefixed tothe product 84, as there are 
two places of decimals in one factor, and one in the other. The 
product, therefore, is .084. 


Notes. — Recourse should be had to our short methods, whenever possible, in 
multiplying decimals, as well as whole numbers, as illustrated in the following: 


Exam. 3. Multiply 79.96 by 79.94. 


Treating the factors, in this, as whole numbers, 4 and 
6 make 10, and the other figures are alike: say 4 times 79.96 
6 are 24; set down in full; then, add 1 to 799 making 79.94 
500, and say 800 times 799 are 63920 to complete the 639.2024 
product. Pointing off four places now, we have 
633.2024. (See page 87.) 


258 Hints AND HELPS FoR THE STUDENT. 


The student’s attention is called to the remaining pages of the 


work, which are devoted exclusively to short methods. These 


methods, so far as known to the author, are entirely original, and 


will be found both interesting and practical : 


To Mu.ttiety THE ’TrEns ToGETHER. 


Rue. (1) Multiply the units by the units and set down the unit 
figure of the product. (2) Add the figure to be carried, if any, to 
either factor, and to the result; add the unit figure of the other 
factor, dropping the 1 from that other factor, always. 


Nore. —It is immaterial to whick factor the carried figure is added, provided 
the 1 is dropped from the other, as illustrated in the following: 


EXAMPLES : 
A X16 272: Here, say 6 times 7 are 42; set down 2, and carry 
19 x 18 4 to 16 are 20, and 7 (in 17) are 27: Or, 4 to 17, are 
16 x 143 21, and 6 (in 16) are 27, making 272; the 1 being 
etc. dropped from the opposite factor in either case, 
18 X 154 = 276: In this, say 4 of 18 is 6, to carry: then, 5 times 
174 x 18 8 are 40, and 6 are 46; set down 6, and carry 4 to 
etc. 18, are 22, and 5 (in 15) are 27: Or, 4 to 15, are 19, 


and 8 (in 18) are 27; making 276. The reason for 
dropping the 1 from either factor will be understood from the 
following : 
If the parts of the numbers be multiplied together, 
instead of the numbers themselves, as shown in the 


margin, we have, first, 10 x 10 = 100; then 10X6= 17=10+7 
60; next, 7x 10= 70 and finally, 7x6—=42. Add- 16=10+6. 
ing the several products thus found, we get 272, or sh 
16 times 17. And here, it will be observed that, 70 
having set down 2, the unit figure in 42; we add, or 4/2 
carry 4 to 6 and 10 (16) leaving 7 (the 1 being dropped) O718: 


to be still added. Or, carrying 4 to 7 and 10 (17) 
leaves 6 (the 1 peing dropped) to be added. 
And when the ’teens are taken in a reversed order, 
12, 18, 14, 15, 16, 17, 18 and 19 will become: 
21, 31, 41, 51, 61, 71, 81 and 91. each ending in 1. 


ee ee 


Hints AND HELPS FOR THE STUDENT. 259 


Now, 

Any two figures ending in 1 can be multiplied together by the 
following simple rule: (1) Set down the unit figure. (2) Add the 
tens and set down the unit figure of the sum. (8) Multiply the tens 
together, carrying as usual; thus: 


vi X 81 = 7371; In this, set down the unit figure 1; then add: 


bx. 01 8 and 9 are 17; set down 7, and carry 1: now mul- 
ox. 11 tiply; 8 times 9 are 72, and 1 are 75 completes the 
etc. product. - j 


The reason for adding in the second part of the process 
will be understood if the numbers be multiplied together, 91 


in the usual way, as shown in the margin. Here, we see 81 
that, having set down the unit figure, the figures 8 and 9 Eh 
are repeated in the work, and added, which can be done "371 


without setting them down a second time. 


To Mottipty tHe Twenties, Turrties, Fortizs, Eve., 
ToGETHER. 
The Twenties, Thirties, Forties, Fifties, Sixties, Seventies, Eighties 


and Nineties, when written in the natural order, can be multiplied 
together by the following: 


Rue. (1) Multiply the units by the units. (2) Multiply the sum of 
the units by a single figure of the tens. (3) Multiply the tens by the 
tens; carrying as usual; thus: 


23 X 24 = 592: Here, we say 4 times 3 are 12; set down 2, and 


29 X 27 carry 1: then, 2 times 7 (4+ 8) or, 7 times 2, are 14, 
lee and 1 are 15; 5, and carry 1: next, 2 times 2 are 4, 
ee. and 1 are 5 completes the product. 


The factors in these exampies, are written in their natural order, 
and here it may be remarked that, when thus written, the tens in 
each set of numbers are alike, and the reason of the rule will be 
understood from the following: 


260 Hints anp HeELps FoR THE STUDENT. 


Exam. Multiply 73 by 72, as shown in the margin: 


Here, having set down 2 times 8, we have next, 2 times 7; 73 
then, 7 times 8, or 3 times 7; that is, 5 (2 + 8) times 7; and TR 
finally, 7 times 7, plus the figure carried. 5256 
37 X 38 = 140€: Say 8 times 7 are 56; 6 and carry 5: now 8 times 
32 x 36 15 (8+ 7) are 45, and 5 are 50; 0, and carry 35: 

etc. then, 3 times 3 are 9, and 5 are 14; making 1406. 
43 x 48 It will be seen that the rule is applicable to all the 
42 x 45 numbers down the margin; but in all cases where the 

etc. unit figures equal 10, while the tens are alike, the process 


62 x 69 can be still further simplified by the following Rule: 
67 x 64 (1) Multiply the units.together and set down the product 
etc. in full. (2) Add 1 to either of the tens and multiply the 
other by the figure thus increased ; as: 
%6 X 74 = 5624: In this, the units, 4 and 6, make 10, and the 7’s 
48 X 72 are alike. Say 4 times 6 are 24; set down in full; 
etc. now, add 1 to 7, and say 8 times 7 are 56 to com- 
plete the product. 
81 X 89 = 7209: When the unit figures are 1 and 9 they give 
61 x 69 only one figure when multiplied, and in such 
etc. cases a cipher is set in the second place, as 
81 x 89 = 7209. (See page 87, and note.) 
The rule is applicable to numbers of three figures, also; and in many 
cases, to four, or more figures; thus: 


126 x 124 = 15624: Say 4 times 6 are 24; set down in full; add 
123 X 127 1 to 12 and say 12 times 13 are 156; making 
15624. 
398 x 392 = 158016: Say 2 times 8 are 16; add 1 to 89, and say 40 
493 x 497 times 39 are 1560; making 156016. 
7994 x 7996=63920024: Say 6 times 4 are 24; set down in full; 
3998 xX 3992 then, add 1 to 799 and 799 x 800 = 639200 
etc. completes the product. 


And the rule can be extended to all numbers of a like nature. 

Having thus shown how the twenties, thirties, etc., can be mul- 
tiplied together when written in the natural order, let us take them, 
now, in a reversed order: Thus, reversing 


23, 24; 382, 37; 48, 49; 57, 58; 73, 76; 87, 89; 92, 98, etc., we have 


Hints AND HELPS FOR THE STUDENT. 20s 


82, 42; 23, 73; 84, 94; 75, 85; 37, 67; 78, 98; 29, 89, etc., and here 
it will be observed that the wnits, in each set of numbers, are alike. 

Now, 

To multiply any two figures by any other two, whose units are alike, 
we give the following Rule: (1) Multiply the units by the units. 
(2) Multiply the sum of the tens by a single figure of the units. 
(3) Multiply the tens by the tens; carrying a3 usual; thus: 

32 X 42 = 1344: Say 2 times 2 are 4; set down 4; then, 7 (4+ 3) 
53 X 48 times 2 are 14; 4 and carry 1: next, 4 times 3 are 
ov X 67 12, and 1 are 13 completes the product. 


The reason is plain: Take 74 x 64, as shown in the margin, 
74 xX 64 = 4736: Having multiplied the units together, 


83 x 93 and set down 6, we have 4 times 7; then, 14 
94 x 74 6 times 4, or 4 times 6; that is, 4 times 13 hen: 
etc. (6 + 7) plus the 1 carried; then 6 times 7; 4736 
plus 5 


And the rule can be applied with equal facility to numbers of 
three figures, whose tens and hundreds consist of the ’teens and whose 
units are alike; thus: 


174 XK 164 = 28536: In this, we say 4 times 4 are 16, 6, and carry 1: 


163 x 183 then, 4 times 33 (16-+17) are 182, and 1 are 133; 
148 x 158 set down 38, and carry 18: finally 16 times 17 
123 x 133 (short method, page 258) are 272, and 13 are 
156 x 156 285, which completes the product. 


Norr. —It will be observed, on examination, that the method made use of in 
multiplying 174 by 164, is the same as that in multiplying 74 by 64. 


And if the three figures be such that the hundreds in both factors 
are alike, while the units and tens consist of the ’teens; as, 417, 416, 
318, 319, etc., we have the following 

Rue. To multiply any two numéers of three figures together, 
whose hundreds are alike, and whose units and tens consist of the 
*teens: (1) Multiply the ’teens together and set down the units and tens 
of the product, carrying the hundreds. (2) Multiply the sum of the 
“teens by a single figure of the hundreds; set down the units and tens 
of the product, and carry the hundreds. (3) Multiply the hundreds 
together to complete the product; thus: 


962 Hints ANnpD HELPs FoR THE STUDENT. 


417 x 416 = 178472: In this, 16 times 17 are 272 (short method); 


115 x 114 set down 72, and carry 2 hundred: then 4 times 

817 x 8138 33 (16 + 17) are 182, and 2 are 134; set down 

215 « 215 34, and carry 1: next, 4 times 4 are 16, and 1 
etc. are 17 completes the product. 


Should the tens in these numbers be a cipher, instead of 1, pro- 
ceed as follows: (1) Multiply the units together and set down the pro- 
duct in full. (2) Multiply one figure of the hundreds by the sum of — 
the units, setting down the product, also, in full. (8) Multiply the 
hundreds together and set down the product in full; thus: 

804 < 803 = 645612: Here, we say 3 times 4 are 12; set down in 


602 x 608 full; then, 7 (8 + 4) times 8 are 56; set down, 
705 < 705 also, in full; and 8 times 8 are 64 completes 
etc. the product. 


In multiplying numbers of this nature together, it is well to 
remember that two figures are always set down for the product of the 
units; and two, also, for the product of the hundreds by the sum of 
the units. When the product of the units is only one figure, a 
cipher is put in the second place; and when the product of the 
hundreds by the sum of the units consists of three figures, two only, 
are set down, the third, or hundreds being carried to the next pro- 
duct; illustrated in the following : | 
803 x 803 = 644809: Here, 8 times 3 are 9 (one figure) set down 


704 x 702 two, 09; then, 6 (8+ 8) times 8 are 48; set 

801 x 809 down in full; next, 8 times 8 are 64 completes 
etc. the work. 

906 « 907 = 821742: Say 7 times 6 are 42; set down in full; then, 

904 x 908 13 (6 + 7) times 9, or 9 times 18 are 117 (three 

807 < 807 figures); set down two, 17, and carry 1 hundred; 

709 « 708: finally, 9 times 9 are 81, and 1 are 82. 


The rule can be applied with equal facility to numbers of four 
figures, also; when the tens in both numbers are ciphers and the third 
and fourth figures consist of like ’teens; thus: 

1103 « 1106 = 1219918: In this, say 6 times 3 are 18, and set 

1204 « 12038 down in full; then, 9 (8+ 6) times 11 are 

1305 « 1804 99; set down in full; next, 11 times 11 are 
etc. 121 completes the work. 


Hints anp Hexps ror THe Srupenr. 263 


Or, commencing with the left hand figures: we have 11 times 11 
are 121; set down in full; then 11 times 9 (8 + 6) are 99; set down 
in full; and 3 times 6 are 18 completes the work as before. 

It may be proper to remark here, also, that in multiplying 
numbers of this nature together, two figures, or a significant figure 


- and a cipher to make two, are always set down for the product of 


the units. And two figures only, are set down for the product of the 
teens by the sum of the units. Should the product give three 
figures, the units and tens are set down, and the hundreds are 
carried to the next product; thus: 


1903 x 1903 = 3621409: Here, 3 times 3 are 9 (one figure) set down 


1804 « 1802 two, 09; then 6 (8+ 3) times 19 are 114 
1701 « 1709 (three figures); set down two, 14, and carry 
1608 x 1601 1 hundred: finally, 19 times 19 (short 

etc. method) are 361, and 1 are 362 completes 


the product, 


And when the foregoing numbers are reversed they can be mul- 
tiplied together with equal facility; thus: 906 and 907 become 609 
and 709, the units in both numbers being alike, while the hundreds 
are unlike: 


609 x 709 = 431781: Here, 9 times 9 are 81; set down in full; 


907 x 807 then, 9 times 13 (6+ 7) are 117 (three figures); 

406 < 706 set down two, 17, and carry 1 hundred; next, 

etc. 7 times 6 are 42, and 1 are 43 completes the 
product. 


And any two numbers of four figures whose units and tens consist 
of the ’teens, the third figure being a cipher, while the fourth in 
each is alike, can be multiplied together by the following: 


RuLE. (1) Multiply the "teens together (short method) and set down 
the product in full. (2) Multiply the sum of the ’teens by one figure 
of the thousands and set down the product, also, nm full. (3) Multiply 
the thousands together to complete the product; thus: 

4017 x 4016 -= 16182272: In this, say 16 times 17 are 272; set 

83014 « 3018 down in full; then, 4 times 33 (16+ 17) are 

5015 x 5015 132; set down, also, in full; next, 4 times 4 
etc. are 16 completes the product. 


264 Hints AND HELrs FoR THE STUDENT. Fra 


Or, commencing with the left hand figures, say 4 times 4 are 16; 
set down in full; then, 4 times 33 are 132; set down in full; and 16 
times 17 are 272 completes the product, as before. 

If, when the sum of the ’teens is multiplied by one figure of he 
thousands, the product gives only two figures, then a cipher is set 
down to make three places, always; thus: 

2019 x 2019 = 4076861: Here, 19 times 19 are 361; set down in 


3014 x 3016 full; then 2 times 38 (19 +19) are 76 (two 
3011 ~« 8015 figures), set down three, 076; next, 2 times 
etc. 2 are 4 completes the product. 


From the foregoing examples and illustrations, the student will 
find no difficulty, now, in extending the rules to the twenties, 
thirties, forties, etc., thus: 

To multiply any two numbers of three figures together when the 
hundreds are alike, and whose units and tens consist of the twenties, 
thirties, ete.: 

RULE. (1) Multiply the twenties, thirties, etc., together and set down 
the two first figures of the product, always, carrying the others. 
(2) Multiply the sum of the twenties, etc., by a single figure of the 
hundreds and set down two figures, only, of the product. (8) Multiply 
the hundreds together, adding the carried figures to complete the 
product. 


Exam. Multiply 823 by 824. 


In this, we say 24 times 23 are 552 (short method, 
page 259) set down 52, and carry 5 hundred: then, 8 8/23 ) AY 
times 47 (23 + 24) are 376, and 5 are 881; set down 81, __ 82d f 
and carry 3 hundred: next, 8 times 8 are 64, and 3 are 675152 
67 completes the product. 


And if ciphers come between the hundreds and the twenties, thirties, 
etc., the process will be still more simple; thus: 


Exam. Multiply 8028 by 8024. 


Here, we have 23 & 24 = 552 (short method) which is 
set down in full; then, 8 times 47 (23 + 24) are 3876, Ba 
which is set down, also, in full; next, 8 times 8 are 64 Ta DRERE 
64376552 
completes the product. 


Hints anp HELps For THE STUDENT. 265 


Exam. Multiply 80023 by 80024. 


In this, we have 23 x 24 = 552; set down in full; ‘ 
then a cipher: now, 8 times 47 are 376; set down in aaa 
full; then a cipher: next, 8 times 8 are 64 completes 6403760559 
the product, 


Exam. Multiply 8094 by 8093. 


In this, we have 93 times 94 (short method) = 8742; 80194 
set down 742, and carry 8: then, 8 times 187 are 1496, 80/93 1 187 
and § are 1504; set down 504, and carry 1: next, 8 65504742 
times 8 are 64, and 1 are 65 completes the product. igh 


Exam. Multiply 80004 by 80093. 


Here, 94 x 98 = 8742; thus: 3 times 4 are 12; 
2 and carry 1: then, 7 (8+ 4) times 9, or 9 times 7 
are 63, and 1 are 64; 4 and carry 6: next, 9 times 9 800/94 l 487 
} 800/93 ) 
are 81, and 6 are 87; set down 8742 in full; now, 8 6414968742 
times 187 are 1496; set down, also, in full; finally, 
8 times 8 are 64 completes the product. 


Exam. Multiply 1034 by 10388. 


In this, 38 times 34 are 1292 (short method); four 
figures, set down only three, 292, and carry 1: then, 10134) , 
72 <1 plus 1 carried are 73 two figures, set down 10 38 7 
three, 073; next, 1 multiplied by 1 completes the . 1073292 — 
product. 


REMARK. — In multiplying numbers of this nature together, it will 
be observed that, when the factors consist of three figures, only two 
are set down for the product of the units and tens; and two figures, 
also, for the product of their sum by the hundreds. When one 
cipher intervenes, three figures are set down in each case, and when 
two ciphers intervene, four figures, and so on, according to the 
number of ciphers. When less than the required number of figures 
in the product, a cipher is set down, as 073, in the last example. 


266 Hints anp HE.LpPs FOR THE STUDENT. 


Exam. Multiply 160042 by 160048. 


Here, 42 x 48 = 2016; set down in full; then, 16 1600/42 t 90 
times 90 are 1440; and 16 times 16 are 256 completes 1600/48 
the product. 25614402016 


The process will be found equally simple tn cases of three figures 
where the hundreds are not alike, the units and tens in both factors 
consisting of any figures; and, also, under similar conditions, when 


ciphers intervene; thus: 


Exam. Multiply 518 by 314. 


In this, 14 times 18 (short method) are 252 (three 
fizures); set down two, only, 52, and carry 2: then, 3 
times 18 are 54, and 2 are 56 and 5 times 14 are 170; 
now, 70 and 56 are 126; set down 26, and carry 1: next, 
3 times 5.are 15, and 1 are 16 completes the product. 


Exam. Multiply 518 by 304. 


Here, 4 times 18 are 72; set down in full; then, 3 
times 18 are 54 and 5 times 4 are 20; the sum of both is 
74; set down in full; now, 3 times 5 are 15 completes 
the product. 


Exam. Multiply 473 by 604. 


In this, 4 times 73 are 292 (three figures); set down ~ 


two, 92, and. carry 2: then, 6 times 73 are 488, and 2 
are 440 and 4 times 4 are 16; the sum of both is 456; 
set down 56, and carry 4: now, 6 times 4 are 24, and 4 
are 28 completes the product. 


Exam. Multiply 4073 by 604. 


Here, 4 times 73 are 292; set down 92, and carry 2: - 
then, 6 times 73 are 488, and 2 are 440 and 4 times 40 
are 160; the sum of both is 600 ; set down in full; 
next, 6 times 4 are 24 completes the product. 


5{18 2 56 
_ 8144 70 
162652 


5/18 ) 54 
3/04 § 20 


157472 


4I73 ) 440 
6045 16 


285692 


ere 440 
6104 


160 
2460092 


Hints anp HEeEtps FoR THE STUDENT. 267 


Exam. Multiply 7034 by 6023. 


In this, 23 times 34 are 782; set down in full; "0184) 204 
then, 6 times 34 are 204 and 7 times 23 are 161; the he a5 161 
sum of both is 865; set down in full; now, 6 times 7 42365782 
are 42 completes the product. 


Before proceeding with our examples, it will be of advantage to 
the student to become thoroughly familiar with the following 
method : 

To multiply any two digits by any other two, in a@ single line; 
illustrated in the following : 


Exam. Multiply 73 by 82. 


In this, say 2 times 73 are 146; set down 6, and carry 73|146 
14: then, 8 times 8 are 24, and 14 are 38; 8 and carry 3: 82 
now, 8 times 7 are 56, and 3 are 59 completes the product. 5986 


Exam. Multiply 67 by 42. 


Here, 2 times 67 are 134; set down 4, and carry 13: then, 67 
4 times 7 are 28, and 13 are 41; 1 and carry 4: now, 4 times 42 
6 are 24, and 4 are 28 completes the product. 2814 


Notgz.— When the student has become familiar with the process, it will not be 
necessary to set down the product in the margin, as shown in the first example, 


And when ciphers intervene, proceed as follows : 


Exam. Multiply 607 by 402. 


In this, 2 times 7 are 14; set down in full; then, mul- 
tiply crosswise: 2 times 6 are 12 and 4 times 7 are 28; ae 
now add both products; 28 and 12 are 40; set down in shot 


full; and 4 times 6 are 24 completes the product. 


Exam. Multiply 7003 by 4006. 


Here, 6 times 3, or 18, is set down, then a cipher to 
make three places; now, 6 times 7 are 42 and 4 times 3 7003 
are 12; 42 and 12 are 54; set down, then a cipher to 4006 
' make three places, also; now, 4 times 7 are 28 completes 28054018 
the product. 


268 Hints AnD HE.LPs FoR THE STUDENT. 


REMARK.—In multiplying numbers of this nature together, it will 
be observed that, when only one cipher intervenes, two figures are. . 
set down for the product of the units; and two figures, also, for the 
result found in multiplying crosswise; and when two ciphers inter- 
vene, three figures are set down in both cases, and so on, according 
to the number of ciphers. . When the products are less, a cipher, or 
ciphers are set down to make the required number, as in the fore- 
going example; illustrated also in the following: 


Exam. 702 by 504. 


In this, set down 4 times 2, then a cipher, 08, to make 702) 10 
two places; then, 4 times 7 are 28 and 5 times 2 are 10; oat 98 
28 and 10 are 38; set down in full; and 5 times 7 are 35 353808 
completes.the product. 


And any number of three digits can be multiplied by any other 
three, in a single line as follows: 


Exam. Multiply 724 by 348. 


Here, by the method for multiplying two digits by sas 83 
any other two, given on the preceding page, we have 3/48) 836 
24 x 48 = 1152; set down 52, and carry 11: then, 3 201952 
times 24 are 72, and 11 are 83 and 7 times 48 are 336, 
both set on the margin to the right; now, the sum of both numbers 
is 419; set down 19, and carry 4: next, 8 times 7 are 21, and 4 are 
25 completes the product. 

If ciphers intervene proceed as follows: 


Exam. Multiply 7024 by 3048. 


In this, 24 X 48 = 1152: set down three figures, 

152, and carry 1: then, 3 times 24 are 72, and 1 are 73 a rete A 
and 7 times 48 are 336; their sum is 409; set down in 51409152 
full; next, 3 times 7 are 21 completes the product. 


Hints Anp Hers FoR THE STUDENT. 269 


GENERAL SHORT METHOD. 


The following short method can be applied in all cases: 


Exam. 1. Multiply 78 by 63. 


When the multiplier consists of two figures: 


- Here, say 3 times 8 are 24; set down 4 for the first 78X63 
figure of the answer one place to the right of the 93 

multiplicand; carry 2, then, 3 times 7 are 21, and 2 are A01k 

23; set under the units and tens of the multiplicand, as 

shown: hext, 6 times 8 are 48, and 3 (add downwards) 

are 51; set down 1 and carry 5; 6 times 7 are 42, and 5 are 47, and 

2 (adding downwards) are 49 completes the product. 


Exam, 2. Multiply 3478 by 74. 


In this, 4 times 3478 are 13912; set down 2 for the 
first figure of the answer, and 1891 under the multi- 3478 X74 
plicand as shown: then, 7 times 8 are 56, and 1 are "7907 hue 
57; set down / and carry 5; 7 times 7 are 49, and 5 557372 
are 54, and 9 (add downwards) are 63. Proceeding, : 
we obtain 257372, the product. 


Exam. 3. Multiply 2356 by 347. 


When the multiplier consists of three figures: 


’ Here, we set 2, the first figure of the answer, two 2356347 
places to the right, and 1649 is set as shown in the “Teo ee 


margin: next, 4 times 6 are 24, and 9 are 33, which 943 
gives the second figure of the answer, and 3 to carry; 817532 


then 4 times 235 plus 3, gives 943, which is set in 

proper position, making the second partial product 9483, the unit 
figure 3 being the second of the answer. Now, 3 times 6 are 18, and 
7 (4+3 adding downwards) are 25; set down 5 and carry 2; 3 times 5 
are 15, and 2 are 17, and 10 (6+4 downwards) are 27. Proceeding, 
we obtain the product, 8175382. 


970 Hints anp Hetps ror THE STUDENT. 


Exam. 4. Multiply 15673 by 5432. 


When the multiplier consists of four figures : 


In this, 6, the first figure of the answer is set three 156735432 
places to the right, and the remain‘ng part of the 3134 
product, 3134, is set as shown: next, 3 times 3 are 9, A702 
and 4 are 13, gives 3 for the second figure of the 6269 
’ answer and 1 to carry; and 4702 set down as shown: 85135736 

then, 4 times 3 are 12, and 5(8+2) are 17, gives 7 
for the third figure of the answer, and 1 to carry: and 6269 is set in 
proper position. Finally 5 times 3 are 15, and 10 (1+0+49) are 25, 
this gives 5, the fourth figure of the answer, and 2 to carry. Pro- 
ceeding thus, we obtain 85135736, the complete product. 


Exam, 5. Multiply 2346 by 3404 


Here, the product by 4 is 9384; 4 is set three places 2346 x 8204 
to the right, for the first figure of the answer,and — 938 
938 as shown: then, the next figure of the multiplier 469 
being a cipher, the 8 is brought down for the second 7516584 
figure of the answer. Now, 2 times 6 are 12, and 3 
are 15, gives 5 for the third figure, and 469 is set in proper position 
as shown. Finally 3 times 6 are 18, and 9+9 are 36, gives the 
fourth figure of the answer, and 3 to carry. Proceeding, we 
obtain 7516584, the complete product. 


Nore.—It will be observed that the unit figure of the answer is set to the right 
of the multiplicand as many places as there are figures in the multiplier less 
one, and that the partial products are one less than the significant figures of the 
multiplier, always. 


Exam. 6. What is the cost of 347 lbs. of sugar, @ 
$5.32 per hundred ? 


UsuaL METHOD. SHORT METHOD. 
347 3475.32 
5.32 69 
694 105 
1041 $18 . 4604 
17385 


$18. 4604 


Hints anp HeELPps FoR THE STUDENT. 271 


SQUARING OF NUMBERS. 


Exam. What is the square of 73 ? 


In this, say 3 times 3 are 9; then, 6 (58 + 3) times 7, or 7 73 6 
times 6 are 42; 2, and carry 4: now, 7 times 7 are 49, and 4 13 
are 538 complete the square, (See pages 259, 260 and 261.) D829 


Exam. What is the square of 417 ? 


Here, the units and tens consist of the ’teens and the 
hundreds are alike: 17 times 17 (short method, page 258) 
are 289; set down 89, and carry 2: then, 4 times 34 ae I 34 
(17 + 17) are 136, and 2 are 138; set down 38, and carry 173889 — 
1: now, 4 times 4 are 16, and {1 are 17; this completes 
the square. 


‘Exam. What is the square of 824 ? 


In this, the units and tens consist of the twenties, 
and the hundreds are alike: 24 times 24 (short method, 
page 259) are 576; set down 76, and carry 5: then, 8 als 
times 48 are 384, and 5 are 389; set down 89, and carry 678976 
53: now, 8 times 8 are 64, and 8 are 67 completes the 
square. 


REMARK.— It may be well to remark, here, that, in squaring 
numbers of three digits, only two figures are set down for the pro- 
duct of the units and tens; and fwo, also, for the product of their 
sum by the hundreds, the remaining figures being carried in both 
cases. 


Exam. What is the square of 1236 ? 


Here, 36 X 36 = 1296; set down 96, and carry 12: 
then, 12 times 72 are 864, and 12 are 876; set down 76, 3 ms 73 
and carry 8: now, 12 times 12 are 144, and 8 are 152 1527696 
completes the square. 


When a cipher intervenes any number of three figures can be 
squared at sight; thus: 


272 Hints AND HELPS FOR THE STUDENT. 


Exam. What is the square of 808 ? 


In this, say 8 times 8 are 64; set down in full; then, 808 
double 64; 2 times 64 are 128; set down 28, and carry 1: 808 
now, 8 times 8 are 64, and 1 are 65 completes the square. 652864 


Exam. What is the square of 12034 ? 


Here, the units and tens consist of the thirties, and 


we have 34 X 34 = 1156 (short method); set nown 156, 120134 68 
and carry 1: then, 12 times 68 are 816, and 1 are 817 120|84 
(three figures) set down in full; now, 12 times 12 are 144817156 


144 completes the square. (See pages 265 and 266.) 


Exam. What is the square of 12734 ? 


In this, we first multiply as if the 7’s were ciphers, 12/7 34 
that is, 12034 x 12034, as in the previous example. 12/7 at 68 
To this we have to add 7 times 68 (84 + 34); 7 times ase 
7 and 7 times 24 (12 + 12); that is, 700 times, 7 being 173376. 
in the hundreds’ place; thus: 7 times 8 are 56; 6 and 162154756 
carry 5: 7 times 6 are 42, and 5 are 47; 7 and carry 4: 

7 times 7 are 49, and 4 are 53; 3, and carry 5: now, 7 times 24 are 
168, and 5 are 173; the sum of both products is the square. (See 
page 270.) 


Exam. What is the square of 70342 ? 
Here, 42 x 42 = 1764; set down in full; then, 7 


times 84 are 588 (only three figures) ; set down four, 703142 
0588; now, 7 times 7 are 49; the result is the pro- 703|42 t 84 
duct of 70042 by 70042, to which is added 8 (that is, 4905881764 — 
300) times 3| 84, or 1152 and 8 times 14 (7+ 7) or 6 421152. . 


(3 + 3) times 7 are 42; the sum of both products is 4947996964 
the square. 

The following will be found interesting and practical: 

Any number of three figures can be multiplied by any other number 
of three figures when the hundreds in both are alike, and whose units 
and tens, when added, make 10, 20, 30, 40, 50, 60, etc., 100, 110, 120, 
1380, 140, etc., as follows: 


| 


Hints AnD HeELPs FOR THE STUDENT. 273 


' Exam. Multiply 742 by 708. 


< In this, 42 and 8 make 50, and the hundreds are alike: 5 
Omit the cipher in 50, and set 5 above the 4: now, say 8 7142 
times 42 are 336; set down in full; then 7 times 75 are 7|08 
525 completes the product. 525336 

Exam. Multiply 518 by 512. 
3 
Here, 12 and 18 make 380, and the hundreds are alike: 5/18 
Say 12 times 18 are 216; and 5 times 53 are 265. 0/12 
265216 


REMARK.— In multiplying numbers of this nature together, three 
Jigures must be always set down for the product of the units and 
tens; when four figures are obtained, the fourth is carried; and 
when only two are obtained, a cipher is set down in the third place; 
illustrated in the following: 


Exam. Multiply 489 by 471. 


In this, 71 and 39 make 110; set 11 above 43: now, 111 
by the short method, page 267; we have 39 X 71 = 2769; A\3 
set down 769, and carry 2: then, 4 times 51 are 204, and att 
2 are 206. 206769 

Exam. Multiply 748 by 702. 

Here, 48 and 2 make 50; set 5 above the 4: now, 2 times 5 
48 are 96 (only two figures); set down three, 096; then, 7 ae 
times 75 are 525 completes the product. 525096 


And when a cipher intervenes in both factors, the process will be 
found equally simple; but in this case four figures are set down for 
the product of the units and tens: if only three are obtained, a 
cipher is set in the fourth place; illustrated in the following: 


Exam. Multiply 3071 by 3089. 


In this, 89 and 71 make 160; set 16 above 07: now, 1/6 
71 x 89 = 6319 (short method); set down in full; then, 3 ot i 
times 316 are 948 completes the product. 9486319 


ite 


974 Hints ano HxExLps FoR THE STUDENT. 


Exam. Multiply 702+ by 7016. 


Here, 16 and 24 make 40; set 4 above the 2. now, 16 Ao 
times 24 are 384 (three figures) set down four, 0384; 70|24 
then, 7 times 704 are 4928 completes the product. Or, 70}16 
having set down 384, say 70 times 704 are 49280. 49280384 


When the units and tens make an even 100; set down their pro- 
duct and make four figures always, setting down a cipher or ciphers 
when four are not obtained; then add 1 to either eu of the 
hundreds, and multiply by che other; thus: 


Exam. Multiply 988 by 912. 


In this, 88 and 12 make 100: say 12 times 88 are 1056; 9/88 
set down in full; now, add 1 to either 9, and say 10 times 9)12 
9 are 90 to complete the product. 901056 


Exam. Multiply 792 by 708. 


Here, 92 and 8 make 100: say 8 times 92 are 736; set 7,92 
this down, then a cipher to make four places; now, 8 708 
times 7 are 56 completes the product. 560736 


Exam. Multiply 1299 by 1201. 
12/99 


In this, 99 times 1 are 99; set this down, then two 12/01 
ciphers; now, 12 times 13 are 156 completes the product. 1560099 


Exam. Multiply 5938 by 5962. 


Here, 62 and 88 make 100, and the hundreds and 
thousands in both factors are alike: now, 62 times 38 59138 
(short method) are 2356; set down in full; then, adding 59/62 
1 to either 59. say 60 times 59 are 3540 to complete the 35402356 
product. | 


And if a figure be changed in either factor, the product is obtained 
with equal facility; thus: 


Hints AnD HE.Lps For THE STUDENT. 275 


Exam. Multiply 5988 by 5964. 


59/38 

In this, we multiply as if 64 were 62, as in the pre- 59/64 
ceding example, and to the result thus found, 2 times 39402356 
5938, or 11876, is added for the required result. 11876 

39414232 
Exam. Multiply 5938 by 5762. 

Here, we assume 57 to be 59, and multiply as in the 59138 
preceding examples, and from the result, 2 (that is, 57/62 
200) times 5938, or 1187600, is deducted; the difference 35402356 
is the required product. And if the multiplier were 11876.. 
6162 instead of 5762, we would add 1187600. 34214756 

Exam. Multiply 15947 by 8953. 

We multiply in this as if 159 were 89, and we have a a 

8947 « 8953 = 80102491, to which is added 7 (that is, 80102491 
62671... 
7000) times 89538, or 62671000, to get...............6. 142773491 


A knowledge of the foregoing methods, with a little practice, will 
now enable us to obtain extraordinary results, without much mental 
effort, in many cases where, by the usual methods, it would require 
considerable labor. Take, for instance, the following: 


Exam. Multiply 120342 by 120358. 


In this, 58 and 42 make 100, and the remaining 1203/42 
figures are alike: 42 x 58 = 2436 (short method, page 1203/58 
267); set down in full; now, adding 1 to 3 (either 14484122436 
one); we have 1204 X 1203: say 3 times 4 are 12; set 
down in full; next, 7 (8 + 4) times 12 are 84; set down in full; and 
12 times 12 are 144; set down in full, completes the product. 
(See page 262.) 


276 Hints anp HELps FoR THE STUDENT. 


Exam. Multiply 82341 by 2359. 
Here, we have, first, 8 (that is, 80000) times 23859, 


or 18872, set five places to the left of units; then, oe 2 
2341 X 2359 = 5522419, which is set in proper posi- eae 
tion, and added; this gives the required product 5599419 


41 x 59 = 2419; then adding 1 to 23, ‘we have 94242419 
24 X 23 = 582. 


7854 


The following method for multiplying any number by 7854, or 
.7854, so much used in practical mathematics, will be found prefer- 
able to the usual method: multiply 84628 by .7854, 

First, multiply by 7 and set down 
the result a second time, one place 34628 ..7854. 


farther to the right; then, double 949396 be 
the latter number and set down the 242396 7854=5 4 40 
result a second time, also, each one 484792 14 


484792 


lace to the right, and add the re. 5" 
place to the right, and a ee” 37196. 8312 


sults. The veason is shown on the 
margin. 


7854 


Numbers ending in 5, when not too large, can be readily multiplied 
together, or squared, by the following: 

Rue. (1) Multiply the units together and set down the product in. 
full. (2) Multiply the remaining figures together and to their product 
add half their sum; thus: 


165 x 45 = 7425: In this, say 5 times 5 are 25; set down in full; 


245 x 65 then, 4 times 16 are 64, and 8 (half of 16) are 72, 
145 x 85 and 2 (half of 4) are 74 completes the product. 
etc. When the sum of the numbers is odd, drop 1 to 


make the number even, then add half this even 
number, and set down 75, always, for the first two figures, instead 
of 25; thus: 
175 & 45 = 7875: Here, the sum of 17 and 4, or 21, is odd, and 
195 x 65 75 is set down, instead of 25; now, 4 times 17 are 
135 x 85 68, and 10 (half of 20) are 78. Or, add 8 (half 16) 
etc, and 2 (half of 4) as you proceed. 


Hints anp HELPs FoR THE STUDENT. OTT 


Now, the square of 695, or 
695 x 695 = 483025: For squaring numbers ending in 5, the rule 


195 x 195 given on page 260 is preferable, since the units 
230 X 285 equal 10, and the other figures are alike always. 
etc. Here, we say 5 times 5 are 25; set down in 


full; then adding 1 to 69, we say 70 times 69 
are 4830 to complete the work. 


The reason for adding half the sum will be understood if we take 
any two numbers ending in 5, say 85 and 45, and multiply their 
parts together, as shown in the margin, 

Here, it will be seen that the partial products, 
when added, make 4825, or 85 x 45, and that, 80 + 5 
omitting the ciphers, we have 4 times 8, or 382, =i Dae 
plas 6 (the half of 4+8) plus 5 times. Anda 00+ 30 
similar mode of reasoning will show why 75 must 3200 a0 : 35 
be set down, instead of 25, when the sum is odd, 


The rule given on page 260, for multiplying together numbers of 
two or three figures whose units equal 10, the otber figures being 
alike, can be applied to mixed numbers whose fractions make 1, and 
whose whole numbers are alike; thus: 

84 X< 84 = 724: In this, the two fractions, when added, make 1, 


Tk X TF and the 8’s are alike: say half by half is a quarter 
6 X 64 (4); then, add 1 to 8 and say 9 times 8 are 72. 
etc. (8;.= 8 5, and 8 5 X 8.5 = 72.25, or 724.) 


Rue. (1) Multiply the fractions together and set down the result. 
(2) Add 1 to either whole number and multiply the other by the number 
thus increased. 


mi62.< 165.— 27245: Here, the sum of 3 and 3 is 1, and the whole 
142 x 144 numbers are alike: now, 3X $=}, which is 
etc. set down; then, 16 times 17 (short method) are 272. 


And any two mixed numbers, each having the fraction 4, can be 
multiplied together by the following Rule: (1) Multiply the fractions 
together and set down the result. (2) Multiply the whole numbers 
together and to their product add hulf their sum; thus: 

OF x 74. = Tt: In this, 4X 4=4, which is set down; then, 7 
183 x 64 times 9 are 63 and 8 (half of 7 + 9) are 71; the pro- 
etc, duct is 71}. ‘ 


278 FRACTIONS. 


When the sum of the whole numbers is odd, add half the next 
lower even number, and set down 2 for the fraction, always, instead 
of +: thus: 

174 & 44 = 783: Here, 17 and 4 are 21, an odd number, set 
194 x 64 down 3; now, 4 times 17 are 68, and 10 (half of 20) 
etc, are 78, 


This is the application of the rule given on page 276, for multi- 
plying together numbers ending in 5. 

GENERAL RULE. To multiply ang two mixed numbers together : 
(1) Multiply the whole numbers together. (2) Multiply the whole 
number of the multiplicand by the fraction of the multiplier. (8) Mul- 
tiply the whole number of the multiplier by the fraction of the multi. 
plicand. (4) Multiply the fractions together; and add the four 
products. 


Exam. Multiply 1434 by 168. 


In this, 16 times 14 (short method) are 224, which is pasadena | 
set down; then, ¢of 14=—103; next, 4 of 16=8 and 104 
4X $= 2: adding the four products, now, we get 242%, 
the required product. Pagar 


The work can be done mentally, in most cases, without setting ~ 
down the several products; thus: multiply 12% by 83. 


Here, we say 8 times 12 are 96, and 4 (4 of 12) are 100; 122 x 84 
and 6 ( of 8) are 106; then, } x 4= ,3, which gives... ot LOG ie 


Now, in actual business transactions, where the fraction of a cent 
cannot be taken (the rule being to reject the fraction when less than 
half a cent, and when a half, or more, to add a cent) the following 
simple method will answer all practical purposes: take 144 lbs. of 
tea, @ 16% ¢ per lb. 


In this, we say # of 14, to the nearest unit, is 11; and 11 8 
3 of 16 is 8; then, 16 times 14 are 224, and 19 (11 + 8) 1434 x 163 
are 248, or $2.48. $2.48 


The correct answer is 2.42%, as found above; but $2.43 for business. 


FRACTIONS. 279 


Exam. What is the cost of 244 yds. of cloth, @ 233 ¢ 
per yard? 


Here, # of 24 to the nearest unit is 18; and the half 18 12 
of 23 to the nearest unit is 12; then 23 times 24 (short 245 X& 23% 
method) are 552, and 80 (18 + 12) are 582. $5.82 


Note. — In multiplying by each fraction, always take what is nearest the true 
result; thus, the 4 of 23 is nearer 12 than 11, etc. 


When either factor is an aliquot part of 100, 1000, etc., or is con- 
veniently near an aliquot part, the multiplication can be more easily 
performed by division. 

The aliquot parts of a number are obtained by dividing it by 
2, 3, 4, 5, 6, etc., as shown in the margin, where the aliquot parts 
of 100 are given. 


RuLE. To multiply by an aliquot part of 100: If 
there be no decimal in the mulliplicand, annex two 
ciphers and divide by 2, 3, 4, etc., as the case may be. 
If there be decimals, or cents, move the decimal 
point two places to the right and divide as directed. 
If a mixed number, reduce the fraction to a decimal. 


© 

oS 
HW UU a ad Tl 
POs Colm aaa Sater saecoptnc|m 


SDD ES? SUP Go 
ae 
i=) 
es|ao 


p= 
— 
Oo 
| 
al 
— 


Exam. What is the cost of 162 yds. of cloth @ $1.25 
per yard? 


Here, 162 being } of 100, instead of multiplying in the 
usual way we move the decimal point two places to the $125 
right in $1.25, this multiplies by 100, and gives $125; $20.83 
now, + of this gives $20.83 the required cost. 


Or, since 125 is 1 of 1000, by reducing the fraction in 16} to a 
decimal, we have 16.666’ to be multiplied by 125. Now, moving 
the point three places to the right, we have 16666. cents, and 3 of 
this gives $20.83 as before. 


280 FRACTIONS. 


Computing THE Cost oF COMMODITIES. 


In computing the cost of commodities, such as dry goods, etc., 
when the commodity contains a fractional part, the process can be 
simplified by reducing the fraction toa decimal and applying the 
method of aliquot parts. 

Take for example 4883 yards of goods at any particular price. By 
reducing 3 to the decimal of a yard, we have 4883 yds. = 488.75 yds. 


Now, the cost of 488.75 yards at $1 per yard = $438.75 
and at 10 cents per yard, the cost is one-tenth of that = 43.875 
at 1 cent per yard, the cost is one-tenth of thatat10cents= 4.3875 
and at 3 cent, the cost is half that at 1 cent = 2.1937 


And font this basis the cost at any given price can be easily ob- 
tained, illustrated in the following examples : 


Exam. 1. Find the cost of 4383 yds. of silk at $1.37} 


per yd. 

In this 4883 yds. =438.75, and at $1 per yd. the cost is $438.75 
Now, 374c.= 25c., plus 124c.; and 25c.=40f $1= .... 109.69 
and at 12ic. the cost is one-half that at 25c.= .... 54.84 
making the total cost $603.28. $603.28 


Exam. 2. What is the cost of 86% yds. at $1.384 per yd. 
Here we have 863 yds. = 86.875 yds. and proceeding as $86 .875 


in the foregoing example, we obtain the three first items 21.718 
which give the cost at $1.374; and the cost at 1c. is 10.859 
found from the first item ($86.875) by moving the point . 868 


two places to the left; this gives .868 making the cost $120.32 
$120.32. 


Exam. 3. What is the cost of 67% yds. at $1.874 per yd.? 


In this, 675 yds.= 67.625; and at $1, the cost= .... $67 .625 
Now, the difference between $1.874 and $2, is 124c., so 185.25 
we multiply by 2 to get the cost at $2 per yd. and 8.453 
from this we deduct one-eighth of the cost at $1, or $126.797 
$8 .453, the cost at 124c. (4 of. $67.625 = $8.453) to get 
the cost at $1.874. 


Note.— To reduce the fractional part to a docintal annex ciphers to the 
numerator, or suppose them annexed, and divide by the ‘denominator. 


ADDITION. 


The following method for Addition requires no “ ¢ar- 
rying,’ and will be of advantage to accountants when 
liable to be interrupted in their calculations : 


3829 . 25 
768 .50 
4687.49 
2823 .35 
7547 .28 
3760.82 
675 . 64 
1846 .35 
3785.10 
23270 .48 
6453.3 
29723 .78 


Commencing at the top of the left-hand column and running 
downwards, we find the sum to be 23, which is set down in full, 
in the usual manner; then, without carrying, we find the sum of 
the next column to be 62; 2 is set in its proper place under the 
column, and 6 to the left, one line lower; the sum of the next 
coiumn, without carrying, is 47; the next 50; then 34, and finaily 
38. Thesum of the two results thus obtained is the required sum. 


Nortg. —It is preferable to write the numbers of each sum in the natural 
order, thus; 62, write down 6 first and then 2, etc. 


289 7 ADDITION. 


Instead of taking one column into consideration, as in 
the foregoing, let us take two columns, as in the follow- 
ing example: 7633.42 

3893.87 
6324.73 


39250. 78 
5 6 


ee 


39756 .78 


Commencing at the bottom of the left-hand column and run- 
ning up, we find the sum to be 34; call this 340; now run down 
the next column, starting with 40, bearing in mind the 8 (300), by 
taking hold of the third finger of the left hand; the sum of the 
two columns is 392, which is set down in full. Next, take the 
third column from the left, without carrying, the sum is 49, call 
this 490 and run down the fourth column starting with 90; tue 
sum of both columns is 550, which is set down in full, 50 in 
proper position, and 5, that is 500, to the left and one line lower; 
the sum of the next two columns is 678, which is set down as 


shown, and the total sum is 39756.78. 
Note.-- If the number of columns be odd, first add the left hand column, 
then two columns at a time, thus: 764.54 
The left-hand column is 26, the sum of the two next, 123.45 
298; and the sum of the two last, 324. 876.43 


23 


Hints AND HeE.Lps For THE STUDENT. 283 


The following rule for Subtraction will be found simple and 
practical : RULE. — Add to the minuend, first, what the untt figure of 
the subtrahend wants of being 10; and for the succeeding figures, add 
what each wants of being 9. Drop 1 immediately to the left of the 
last figure of the subtrahend always. 


Exam. 1. From 74721 take 39864. 


Here, we say 6 and 1 are 7; 3 and 2 are 5; 1 and 7 are 8; T4721 
4is4; 6 and 7 are 13; drop the 1, and the remainder is 39864 
34857. (7)84857 


Exam. 2. From the sum of $687463 and $2346; take 


$6942. 
$687463 + $2346—$6942=$682867. 


In this, we say 8 and 6 are 14, and 3 are 17; 7 and 1 to carry: 
1 and 5 are 6, and 4 are 10, and 6 are 16; 6 and 1 tocarry; 1 and 8 are 
4 and 4 are 8; 3 and 2are 5, and 7 are 12; 2. and 1 to carry; 1 and 8 
are 9; but dropping 1 immediately to the left of the subtractive 
number, we set down 8; then setting down the last figure, 6, the 
remainder is $682867. 


Exam. 3. Received $756.47; $982.34; $765.26; and 
paid out $476.29; what is the balance on hand? 


In this example, we set the subtractive number under $756 .47 
those to be added, and perform the whole process by 982.34 
addition, adding mentally, what the first figure of 765. 26 


$476 29 wants of being 10; and what each remaining Cesanaron 
476.29 
figure wants of being 9. The sum of the last column git oS 
is 30; but dropping 1 from 3, immediately to the left of $2027.78 
$476.29, makes it 2. 
The reason of dropping 1 to the left of the last figure of the sub. 
trahend, always, will be understood from the following : 


984 Hints AND HeELps FoR THE STUDENT. 


Any number, which, when added to another, makes 10, 100, 1000, 
10000, etc., is called the complement of that other. Thus, 434+57= 
100; 43 is the complement of 57; and 57 is the complement of 43. 
The complement of $476.29 is $523.71, both numbers, when added, 
making $1000. 

Subtracting either number, now, from 1000, is the same as adding. 
its complement and dropping the 1000. 

Thus, subtracting $476.29 from $1000 
leaves $523.71; eee as adding $523.71 $1000 $1000 
and subtracting $1000, or dropping the 1 in 416.29 ek 
1000. $523.71 ()528.71 

And this being understood, any number can be subtracted from 
another, by adding (mentally) the complement of the number to be 
subtracted, dropping 1 to the left of that number, always. (See 
page 221.) 


Exam. 1. From 547632 take 876. 


In this, we say, 4 and 2 are 6; 2 and 8 are 5; 1 and 6 547632 
76 


546756. 


are 7; here we drop 1, and set down 6; then 4, and 5. 


Exam. 2. From 370234 take 18547. 


Here, we say, 3 and 4are 7; 5 and 8 are 8; 4 and 2are 6; 3702384 
1; 8 and 7 are 15; carry 1 to 3 is 4, but here 1 is to be 18547 
dropped, and we set down 3, 851687 


Prosi~ems With TuHerr So.vutions. 985 


The following problems, with their solutions, are given for the 
purpose of showing the student how they and similar problems may 
be solved. 

PAPER PROBLEMS. 


Prop. 1. What is the cost of 187 sheets 25 lb. paper, 
500 sheets to the ream, @ 18c per lb.? 


For the solution of this an similar problems, we give the follow- 
ing simple 

RuLE.— Multiply the number of sheets by twice the weight of the 
ream, and the result by the price ; .or by twice the price, and the result 
by the weight, whichever is most convenient, and point off five decimal 
places, always. 

Solution. —187 X 50 X 18 = $1.68300; or, 187 XK 25 x 36 = 
$1.68300; for business $1.68. 

The reason for pointing off five places: If we proceed in the usual 
way we multiply the number of sheets by the weight and the result 
by the price, and divide by 500. 

Now, by doubling the weight, or the price, we double 500, also, 
making 1000 for the divisor. Three places are cut off for the 
ciphers in 1000, and two for .18 in the multiplication, making five 
places in all. 


Note. —If the ream consist of 480 sheets, proceed according to the rule, and 

at to the result four per cent. of itself (.04) thus, $1.68300 x .04 = .06732; then, 
1 683 
067 


$1.75 
Reason: The difference between 500 and 480 = 20, and when 
ue divide by 500 instead of 480, we obtain a result too small by go's or 


rfp-= .04. 


Pros. 2. 28 lb. paper is listed to be sold at 10c. per 
lb. by the ream, but for a quantity purchased less than a 
ream, 25% extra is charged; what is the cost of 264 sheets 
on these terms ? 

Solution.— 264 Xx 28 K 20 = $1.47840; then 25% = + and we 


have $1.47840 plus 
4=  .36960 


$1.84800 


r, 

By adding the percentage to the number of sheets, the weight, or 
the price, whichever is most convenient, we obtain the same result, 
thus, 264 plus + or 66 = 330, and we have 330 x 28x 20 = 
$1.84800; or, by adding 25% to 28 making it 35, we have 264 x 35 X 
20 = $1.84800. 


286 Proptems Wiru THEIR SOLUTIONS. 


Paper PROBLEMS. 


Weights of paper equal in thickness to 24 x 36. 


Rue. — To find the weights of paper equal in thickness to 24 x 86: 
(1.) Draw a vertical line, and on the left of said line, set the given. 
dimensions ; and on the right set the required dimensions, and the 
given weight. (2.) Divide the product of the numbers on the right by 
the product of those on the left; the result is the required weight. 


Exam. If 24x36 weigh 70 pounds; what will 40x 48 
weigh ? 


aE NS 
Solution. — 24|40 It is scarcely necessary to re- ue 
36/48 mark that recourse may be had to 38 
70 Beep ec in eo As 
—_——————- such cases, thereby shortening the — 
641184400(165.5 process. thuss . lai 


155.5 

Dividing 24 and 48 each, by 24, we obtain 1 and 2; and dividing 
36 and 40 each, by 4, we obtain 9 and 10. We have now 10 x 2 X 
70 +9 = 155.5. 

PrrcenracE PROBLEM. 

If. $30,000 be invested in property which rents for 
$250 per month, and on which $750 are paid in taxes; 
what rate of interest does the investment pay ? 

Solution. — Here, $250 per mo. is $3,000 per year, out of which 
$750 are paid in taxes leaving $2,250 income on the investment. 

Now, if $30,000 in one year give $2,200 income, what income (rate) 
ought $100 give? 

And for this we have the following proportion: 

As $30,000 : $100 : : $2,250 :x, or the rate on $100; and we have 

$225,000-+$30,000—747%. (See Useful Rules, pages 155 to 158.) 
* RuLE. — Divide 100 times the income by the investment. 
Tae Porunation PRospiem. 


If the population of Albany was 90,000 in the year. 
1900, and 100,000 in 1910; what was the rate per cent 
increase during the interval ? 


Solution. — Here 100,000 — 90,000 = 10,000, the increase of popu- 
lation during the interval. 

And we have: As 90,000 :100 :: 10,000 :x = 1,000,000 + 90,000 = 
115%. (See page 158 ) 

RULE. — Divide 100 times the increase of population during the 
interval, by the population of the earlier date; the result is the rate ‘per 
cent. 


PRoBLEMS witH THEIR SOLUTIONS. 287 


Tue Grant PROBLEM. 


General Grant sold 2 horses for $198 each, he gained 
10% on one and lost 10% on the other, what was the 
result of the sale? 


Solution.— In this, rule 111 of Profit and Loss, page 144, is appli- 
cable: To find the first cost from the gain per cent., and the selling 
price. . 

Now, what cost $100 was sold for $110 and we have to find what 
the horse cost which was sold for $198. 

As the selling price, $110 : $100 cost, : : $198 selling price to x, or 
the cost, and we have 100 x 198 + 110 = $180, the first cost. 

Selling price $198, cost price $180; the result was a gain of $18 in 
the first place. 

Next, what cost $100 was sold for $90; and we have the following 
As $90 : $100: : $198 : x = 19800 + 90 = $220, the cost in the second 
place, showing a loss of $22; then $22 — $18 = $4 loss, the result of 
the sale. (See problem 2, page 145.) 


Tue Sauipe PrRoBuem. 


A ship springs a leak 40 miles from shore, and admits 
3% tons of water in 12 minutes; 60 tons will sink her, 
but the pumps throw out 12 tons per hour. Find the 
average rate of speed to bring her to shore before sinking. 


Solution.— The ship admits 33 tons in 12 min., she will admit 5 
times that in 60 min., or 183 tons per hour (83X5=183 tons) but the 
pumps throw out 12 tons per hour: 183 — 12 = 63 tons, the quantity 
admitted per hour. Now, if 63 tons be admitted in 1 hour, how long 
will it take to admit 60 tons? 

The proportion is as follows: 

As 63:60: :1:x=60 + 63 = 88 hours. 

Next, if in 83 hours the ship has to go 40 miles, what must be the 

speed per hour? 
As 8§h.;:1h.: :40m.:x= 40 + 88 = 4} miles, 
Proof: 83 x 43 = 40 miles. 


288 PROBLEMS witH THEIR SOLUTIONS. 


Tue Ort PROBLEM. 


A quantity of flax seed being converted into oil, the 
result was found to be 658 lbs. of oil and 1276 lbs. of 
cake; how much oil is that to the bushel; what per cent. ; 
and what is the value of the oil at 10¢ per gallon; a 
bushel of seed being 56 lbs., and a gallon of oil 74 Ibs.? 


Solution. — Here, the rule given on page 149 is applicable, thus, 
658 + 1276 = 1934, the whole number of pounds obtained from the 
seed. 

Now, as 1934 : 658: : 56: x; and we have 658 x 56+1936=19.05 lbs. 
to the bushel. . 

Erne this by 74 lbs. to the gal. we have 19.05 + 71 = 2.54 gal. 

24 gal. nearly to the bushel. 

‘eed we have 1934: 658: :100:x, or the percentage, that is, we 
have 65800 + 1934 = 84¢ nearly. 

Finally, dividing the number of pounds of oil, viz., 658 by 7% 
gives the number of gallons; 658 + 73 = 87.7 gals. nearly, and at 
10¢ per gal. the value is $8.77. 


Nots. — To divide by 74, add one-third and take one-tenth. (See exam. 1, p. 
249.) 


THe CHICKEN PROBLEM. 


Do figures lie? Let us see. 


Two women sold 30 chickens each and agreed to divide the proceeds 
equally. One sold her chickens 2 for $1, getting $15 for 80 chickens 
and the other sold hers 3 for $1, getting etd, for her 30 chickens. 
This made $25 for 60 chickens. 

The merchant being asked to divide the money, said: You sold 
yours 2 for $1; and yours, 3 for $1, that is 5 for $2; well 5 into 60, 
12 times and 12 times 2 are $24; $12 each. 

But the women received $25; how can this be explained? 

Solution. — 2 for $1 = 1 for 50¢ 

3 for $1 =1 for 33k¢ 
2 for 834¢ or 1 for 413¢ average price, 
Now, 60 @ 412¢ = $25. 


PROBLEMS WITH THEIR SOLUTIONS. 289 


Tuer Contract PROBLEM. 


Four men contracted to do a certain job of work for 
$8600; the first employed 28 men 20da., 10 h. a day; 
the second, 25 men 15 da.,-12 h. a day; the third, is 
men 25 da.. 11 h. a day; and the fourth, 15 men 24 da., 
8 h.aday. How much should each contractor receive ? 


Solution. — 28x 20 10=5600, the number of hrs. for 1st contractor. 


25X15 12=4500, =‘ Py ‘fone oad 
exe 4050, 4 a % ye aard as 
15X24x 8=2880, ‘“ i 4th ss 


17930, total number of hours for $8600. 
And here the rule given on page 149, for division into proportional 
parts is applicable; the question resolving itself into this: 
_ If $8600 be paid for 17930 hours’ of work; what should be paid 
for 5600 h., 4500 h., 4950 h., and 2880h.? For this we have the 
following: 


1st. As 17930 : 5600: :8600 : a ENS = $2686.00 
2nd. As 17930 : 4500: : 8600 : Se = 2158.39 
8rd. As 17930 : 4950 ; : 8600 : age eee = 2374.24 
4th. As 17980: 288: :8600:x esi 1381.37 

$8600.00 


Tur Coat PRoBLeEm. 


A boy agreed to work for a mechanic 20 weeks, on 
condition that he should receive $20 and acoat. At the 
end of 12 weeks the boy quit work and received $9 and 
the coat. What was the value of the coat ? 


Solution. — The boy’s loss in 8 weeks (20—12) was $11 (20—9). 
If he had worked the 8 weeks he should have received $11 and $9. 
Now, if for 8 weeks he should have received $11; what should he 


receive for 12 weeks? 


KaG 1olers hee r= At = $16.50 


he should have received $16.50; but he received ee 9.00 
$9, the coat, therefore, was worth the difference $7.50. £7.50 


290 PropieMs witH THEIR So.urions. 


MarxinG Goops. 


Short methods for trade discounts have already been 
given from page 232 to 237. In connection therewith, 
it may be well to give here.a few examples on the mark- 
ing of goods, calling attention to the fact that losses may 
be sustained when supposed profits are being made. 


Exam. 1. Suppose goods are marked to sell at 40% 
above cost, and we offer a discount of 20%, our profit is 
not 20% but 12%; for the reason the discount is not cal- 
culated on the vost of the goods but upon the marked or 
asking price, which includes the first cost and the per 
cent. of profit. 

Solution.— What cost $100, we have marked at $140 
and on this we allow a discount of 20%, or }...... 28 
showing a profit of only 12¢. $112 

Again, suppose, without full consideration, we offered a discount 
of 30% on the foregoing, instead of 20%; what would be the result? 


Here; what cost $100, we have marked to sell for............ $140 
and 380% of this is found to be $42, which on being deducted... 42 
shows a loss of $2. ; $98 


RuLE.— To mark goods so as to allow a discount on the asking price 
and have them net a certain profit: 
Divide the net, or desired price by 100 less the discount. 


Exam. 2. What must be the asking price of goods, to 
allow the purchaser a discount of 20% and net the manu- 
facturer $25 per dozen ? 

Solution. —Here, we have 100%—20%=802; decimally expressed, .80. 
Dividing the desired price, $25, by .80, we have 2500 + 80 = $31.25. 
Or, 
Applying the rule given under the head of percentage Case V, 
page 118; the question may be asked thus: What number diminished 
by 202 of itself is equal to 25? And for this we have the following 
proportion : } 

As 80:100: : 25: x = 2500 + 80 = $31.25 
Proof: 20¢=iand } of $31.25= 6.25 
deducting $6.25 we obtain. ..... $25.00 


Marxina Goons. 291 


Exam. 3. What must be the asking price of a piano 
costing $350 to allow a discount of 25% to the buyer and 
still gain 20% above cost ? 


Solution.—To realize the desired profit in this, 202, or } of $350.00 
the cost, is added to itself; this gives $420, or the net price 70.00 
at which the manufacturer wishes to sell the instrument. $420.00 

Now, what sum diminished by 252 of itself is equal to $420? 

As $75 : $100: : $420 : x = $42000 + 75 = $560.00 the asking price. 

Hence the following : 

RuLE.— Add the desired profit, or rate per cent. to the cost priee, 
multiply this by 100 (annexing two ciphers) and divide the result by 
100 less the rate of discount. 

Proof: Marked price $560 
Less 25% discount 140 

Net selling price $420 
Less added profit 70 
Cost $350 

RULE.— 70 find what rate per cent. of discount may be allowed on 
the asking price to realize cost: Subtract the cost from the asking price, 
annex two ciphers, and divide the result by the asking price. 


Exam. 4. If the cost of a piano be $350 and the ask- 
ing price $560; what rate per cent. of discount can be 
offered to realize cost ? 


Solution. — $560 — $350 = $210; then $21000 + 560 = 3744¢. 

Case 2 of percentage, page 115, is applicable here: Given the per 
centage and base to find the rate. 

Deducting the cost from the asking price gives the percentage. 
In the present example, 560 is the base and 210 the percentage; and 
the problem may be resolved into this: If the percentage on 560 be 
210; what is the percentage on 100? And for this we have the 
following proportion : } 

As 560 :210: :100:x = 21000 + 560 = 38734. 

_ This explains the foregoing rule. 
Proof: 374% of 560 = $210 and $560 
Less 210 
Cost $350 _ 


999 Marxina Goobs. 


RULE.— To tell quickly what a single article should sell for, when 
commodities are bought by the dozen, to gain a certain per cent. of 
profit: Divide the cost per dozen by 10; the result will be the selling 
price including a profit of 20% always; and from this any per cent. of 
profit may be readily obtained, illustrated in the following examples: 


Exam. 1. If hats be bought for $35 a dozen; what 
must be the selling price of a single hat to make a profit 
of 20%? Ans. 1, of $385 = $3.50. 


The reason for dividing by 10 to obtain a profit of 202, will be 
understood from the following proportion: 

As $100 considered as cost, is to $120, selling price, so is $385 cost 
to its corresponding selling price; and as 12 articles is to 1, so is the 
cost of 12 to the selling price of 1; thus: 

Reducing thistoasimple As100:120: :385:x 


proportion, and cancelling 12 ee 
1200 and 120, we obtain 1200 : 120: :35:x 
10 to 1 as 85 tox, and divid- 10: 1: :35=385+ 10= $3.50 


ing 85 by 10, we get the selling price of a single hat including a 
profit of 20%. (See exam. 3, page 146; and note page 147.) 


Exam. 2. If shoes be bought for $32 a dozen pairs; 
what should be the selling price of one pair to realize a 
profit of 50%? 


Solution.—To make a profit of 20%, we take one-tenth of $32=$3.20 
and this represents 120%; but we desire a profit of 50%, ora 
selling price of 150%; the difference is 30% (150—120) and 30% 
is one-fourth of 120%, so we add to $3.20 one-fourth of itself, .80 
or 80¢; this gives the selling price of a pair, $4. $400 

Proof: 12 pairs cost $82.00 
Added profit, 50%.... 16.00 
Selling price of 1 pair $48.00 + 12 = $4 

Nots.— To make 60% profit, add % of itself to the 20% (160—120=40) and 40% 
is 4% of 120%; to make 30%, add og (120— 120 = 10) and 10% is 149 of 120%; for. 
8314%, add (13314 — 120 = 184) and 1314% is 1% of 128%, ete. And when the 
desired profit is less than 20%, deduct, thus: To make 1624%, deduct 14, (120— 


11634=314) and 314 is 146 of 120%; to make 15%, deduct 14 (120—115=5) and 5% 
is 144 of 120%, etc. 


tING OME UL TLON. 


Involution is the process of raising a number to any 
proposed power. 


A Power of a number is either the number itself, or the product 
arising from using the number a certain number of times as a factor. 

The first power of a number is the number itself. Thus, the first 
power of 5 is 5. When the number is used twice as factor, the 
result is the second power, or the square, of that number; when 
three times, the third power or cube; when four times, the fourth 
power, etc. 

Powers are denoted by a small figure placed above and to the 
right of the number, to show how many times it is taken as a factor; 
thus, 5! = the first power of 5. 

5? = 5 X 5 = 25, the second power, or square of 5. 
5°? =5 xX 5 X 5 = 125, the third power, or cube of 5. 
5¢=5 X5 X5 X 5 = 625, the fourth power of 5. 

The small figure above and to the right of the number is called 
the Index, or the Hxponent, of the power. ; 

Notr.— The number of multiplications is one less than the numbar of factors 


As the first power of a number is the number itself, its index or exponent is 
generally understood. 


PRINCIPLES I. A number is raised to a given power by taking it as 
a factor as many times as there are units in the exponent; thus, 
ei x Xo "1D. 

Il. The product of any two or more powers of the same number, 
is equal to the power indicated by the sum of their exponents thus, 
Pao 419 0) (0 XO KD) = OPT 8 SS BS. 

Nore.—It frequently happens in the multiplication of decimals, or in raising 
a decimal to a proposed power, that a greater number of decimal figures is 
obtained in the product, than is necessary for practical accuracy. This may be 


avoided by making use of the sclcwsne contracted method for the multiplica- 
tion of decimals: 


RULE. (1) Count off, after the decimal point in the multiplicand, 
(annexing ciphers, if requisite) as many figures of decimals as it is 
necessary to have in the product. (2) Below the last of these, write the 
unite figure of the multiplier. and write the other figures in reversed 
order. (3) Then multiply by each figure of the multiplier, thus in- 
verted, neglecting all the figures of the multiplicand to the right of 
that figure, except to find what is to be carried; and let all the partial 
products be so arranged, that their right hand figures may stand in 
the same column. (4) Lastely, from the sum of the partial products, 
cut off the assigned number of decimal places. 


294 | INVOLUTION. 


Exam. Multiply 7.24651 by 81.4632, so that there 
may be three decimal places in the product. 


CONTRACTED METHOD. COMMON METHOD. 
7.24651 Here 1, the unit figure of the multi- 7.24651 
2364.18 plier, is set below 6, the third decimal 81.4632 
579721 ~+figure of the multiplicand; 8, the figure "71449302 
9247 which precedes 1, is put after it; 4, the 94 |73952 
9898 figure which follows it, is set before it, A34'7906 


435 etc. Wethen say, 8 times 5d are 40, and1 9898 604 
99 (carried from 8 times 1) are 41: 1 is then WOABI51 
{ set down ane 4 carried, oo the ces of 517972018 
san ao, the work by 8 proceeds in the usual Way. <a, aan lanai 
090.504 Then in multiplying 7.246 by 1, we add 590.328 893482 
1 tothe product for 51 because 51 is nearer 100 than 0, and therefore 
it is nearer the truth to carry 1 than 0. In multiplying 7.24 by 4, 
3 is carried for the product that would have resulted from the 
rejected figures: for, going two places back, we have 4 times 5 are 
20; 4 times 6 are 24, and 2 are 26, which being nearer 380 than 20, 
we carry 3. 

So likewise in multiplying 7.2 by 6, we carry 3 from the rejected 
figures, and thus we proceed in similar cases. In finding what is to 
_be carried from the rejected figures, it is generally sufficient to go 
one place back, but in doubtful cases it may be well to go farther. 


The reason of the contracted method will be seen from the com- 
mon method, in which a vertical line is drawn, cutting off the part 
rejected in the contracted*method. 


To illustrate the contracted method still further, let us take the 
following 


Exam. Multiply .681472 by .01286, so that the deci- 
mal in the product may contain five figures. 


CONTRACTED METHOD. COMMON METHOD. 
.§814 72 In this example, since the multi- .681472 - 
68210.0 plier contains no integer, a cipher is .01286 
681 placed below the fifth figure of the mul- 4|088832 
186 _ tiplicand; and then, the multiplier being 54/5176 - 
55 written in reversed order, the work 1386/2944 
4° proceeds as in the last example. A 681/472 


.00876 glance at the common method will show __.00876|372992 
the advantage of the contracted method. 
The contracted method will be particularly useful in computations 


in Compound Interest, given on page 296, and illustrated in the 
following. 


INVOLUTION. 


295 


Exam. What will $1 amount to in 10 years at 6% per 


annum, compound interest ? 


1.12360 0 


Solution.— The amount of $1 for one year at 
6% is $1.06, and by raising this to the tenth 6821.1 
power, we get the amount for 10 years. 1 12360 0 

1.06 1.06 = 1.1236 = the second power; 11236 0 
and this multiplied by itself will give the 2947 9 
fourth power (Prin. J7); next, the fourth mul- 937 1 
tiplied by itself will give the eighth power; 67 4 
finally, the eighth multiplied by the second, ETE 
will give the tenth power. ee 7 = 4th 

Note. —In carrying from the rejected figures, we it 74262.1 
should take what is nearest the truth whether it be too 1 26247 7 
great or too small. 95249 5 

In computations in compound interest, the 1574 9 
decimal figures are usuallv carried to six 2029 
places; so we annex two ciphers to 1.1236 so that 50 5 
the product may contain six. Then reversing 88 
1.1236, setting the unit figure under the sixth 9 
figure of the decimal and multiplying by the 1.59384 8 = &th 
contracted method, we obtain the fourth power. 6321.1 — 2d 
Next, reversing these figures, and multiplying —jp9ser8. 
we obtain the eighth power; and finally, multi- 15938 i; 
plying this by the second power reversed, we 31877 
obtain the tenth power, which is the amount of 478 2 
$1 for 10 years at 6%. And from this, the U5 6 
amount of any sum may be obtained by oe 

1.19054 8 = 10th 


multiplication. 


Had the products here been found at full length the work would 


have been quite laborious, requiring 346 figures in the operation, 
and giving 21 figures for the last product, 1.79084769654285362176; 
whereas by the contracted method we obtain 1.790848, six decimal 
figures sufficient for all practical purposes. 

To facilitate the calculation of such problems as the foregoing, 

however, we give on page 298 a table showing at sight the amount 
of $1 at 2% to 10% for any number of years from 1 to 35. 
- Thus, taking the above example, to find the amount of $1 for 
10 yrs. @ 6 per cent. per annum, compound interest; we turn to the 
table and look in the yearly column for 10 years, and opposite in the 
6 % column, we find 1.790848. 


COMPOUND INTEREST. 


Compound Interest is interest on both principal and 
interest, when the interest is not paid when due. 


RuLE.— When the interest is payable annually : 

I. Find the amount of the given principal for one year at the given 
rate, and make it the principal for the second year. 

Il. Hind the amount of this new principal, and make it the prin- 
cipal for the third year, and so continue for the given number of years. 

Ill. Subtract the given principal from the last amount; the re- 
mainder will be the compound interest. 

Notrs.—1. When the interest is payable semi-annually or quarterly, find the 
amount of the given principal for the first interval, and make it the principal 
for the second interval, proceeding in all respects as when the interest is pay- 
able annually. 

2. When the time contains years, months, and days, find the amount for the 


years, upon which compute the interest for the months and days, and add it to 
the last amount before subtracting. 


Exam. What is the compound interest of $800 for 4 
years at 5%? 3 

The amount of ¢1 for 1 year at 54 = $1.05, and this multiplied by 
the given principal will be the amount of said principal for 1 year. 


The solution is as follows: . 
. $800 Principal for 1st year. 


$800 X 1.05 =$840. . “« “ eg « 
840 x 1.05 = $882 6 4 Saas 
$882 X 1.05 = $926.10 « “ Ath « 


$926.10 < 1.05 = $972.40 Amount for 4 years. 
$800 Given principal. 


$172.40 Compound interest. 


If the amount of $1 for 1 year be raised to the power denoted by 
the number of years, the result will be the amount of $1, or £1, for 
the number of years. 

Thus, 1.05 x 1.05 x 1.05 & 1.05 = 1.215506, and this multiplied 
by the given principal gives the amount: 1.215506 x 800 = $972.40, 
as found above. 

As calculations in compound interest are facilitated by the use of 
interest tables, we give on pp. 298 and 299 a table showing the 
amount of $1, or £1 sterling, at compound interest for any number 
of years from 1 to 35. Thus, taking the foregoing problem: Look- 
ing in the yearly column we find 4 yrs. and opposite in the column | 
for 5% we find 1.215506. 


Compounp INTEREST. 297 


Exam. What is the amount of $6000 for 10 years at 
5% compound iaterest, payable semi-annually ? 


Here, the payments being half-yearly, the rate is 24 or .025 and 
the amount of $1 for half a year is 1.025; and since there are 20 
payments, 1.025 is to be raised to the 20th power. To avoid such a 
long operation, we refer to the table to 20 yrs. and 23% and we find 
1.638616, the amount of $1, and multiplying this by the prin. we 
have the required amount, thus, 1.638616 « 6000 = $9831.696. 

Since the t¢me used in expressing any rate of interest is entirely 
arbitrary, and having fixed the ratio between the principal and 
interest at each compounding, the result evidently depends upon 
the number of times the operation is repeated. 

Thus, if the interest be compounded a given number of times by 
adding to each respective amount 9% of itself, it matters not whether 
it be considered 5% per annum or 5% per day, the result would be 
the same. 

Hence, if the intervals be less than a year, as when the interest is 
to be compounded semi-annually or quarterly, the tables constructed 
for yearly intervals can be used by reducing the rate per cent. pro- 
portionally, and taking from the table the proper number of inter- 
vals. If the interest is to be compounded semi-annually, when the 
rate is said to be 5% per annum, 24% should be used at each com- 
pounding, though it would amount to more than 5% compounded 
annually. 


Exam. Required the amount of $5000 for 10 yrs., 
8 mo., 20 da. @ 6% per annum, compound interest, pay- 
able semi-annually. . : 


Solution.—In 10 yrs., 6 mo. there are 21 intervals of six months 
each, leaving 2 mo. and 20 da.; and the rate is 8% semi-annually. 

The amount of $1 at 3% half yearly is 1.03, and this raised to the 
21st power (1.037!) gives the amount of $1 for 21 intervals, or for 
10 yrs., 6 mo. compound interest. Multiplying this amount by the 
given principal, $5000, gives the amount of said principal for 
10 yrs., 6 mo. and adding to this amount its interest for 2 mo., 
20 da. @ 6%, we get the total amount. 

Now, to save time and labor, we refer to the table given on page 
298, and look in the yearly column for 21 years, and opposite, in 
the 3% column, we find 1.860295, the amount of $1, and multiplying 
this by the given principal, we have: 1.860295 x 5000 = $9801.48, 
the compound amount for 10 yrs.,6 mo. Adding to this its interest 
for 2 mo., 20 da. @ 6% = 124.02, we obtain $9425.50, the required 
amount. 


298 Compound Interest TABLE. 


Showing the amount of $1, or £1, compound interest, from 1 to 35 - 
years, at the following rates: - 


2”, Qhe, 34. Big. Ag, 41g 


w 

g 

| 

1 | 1.926000 | 1.025000 | 1.030000 | 1.035000 | 1.040000 | 1.045000 
2 | 1.040400 | 1.050625 | 1.060900 | 1.071225 | 1.081600 | 1.092025 
3 | 1.061208 | 1.076891 | 1.092727 | 1.108718 | 1.124864 | 1.141166 
4 | 1.082482 | 1.108818 | 1.125509 | 1.147523 | 1.169859 | 1.192519 
5 | 1.104081 | 1.181408 | 1.159274 | 1.187686 | 1.216653 | 1.246182 
6 | 1.126162 | 1.159693 | 1.194052 | 1.229255 | 1.265319 | 1.802260 
7 | 1.148686 | 1.188686 | 1.229874 | 1.272279 | 1.815982 | 1.360862 
8 | 1.171659 | 1.218403 | 1.266770 | 1.816809 | 1.868569 | 1.422101 
9 | 1.195098 | 1.248863 | 1.304773 | 1.362897 | 1.423812 | 1.486095 
10 | 1.218994 | 1.280085 | 1.348916 | 1.410599 | 1.480244 | 1.552969 
11 | 1.248374 | 1.812087 | 1.884234 | 1.459970 | 1.539454 | 1.622853 
12 | 1.268242 | 1.844889 | 1.425761 | 1.511069 | 1.601082 | 1.695881 
13 | 1.293607 | 1.878511 | 1.468534 | 1.563956 |.1.665074 | 1.772196 
14 | 1.819479 | 1.412974 | 1.512590 | 1.618695 | 1.731676 | 1.851945 
15 | 1.845868 | 1.448298 | 1.557967 | 1.675849 | 1.800944 | 1.935282 
16 | 1.372786 | 1.484506 | 1.604706 | 1.733986 | 1.872981 | 2.022370 
17 | 1.400241 | 1.521618 | 1.652848 | 1.794676 | 1.947901 | 2.113377 
18 | 1.428246 | 1.559659 | 1.702433 | 1.857489 | 2.025817 | 2.208479 
19 | 1.456811 | 1.598650 | 1.758506 | 1.922501 | 2.106849 | 2.307860 
20 | 1.485947 | 1.638616 | 1.806111 | 1.989789 | 2.191128 | 2.411714 
21 | 1.515666 | 1.679582 | 1.860295 | 2.059431 | 2.278768 | 2.520241 
22 | 1.545980 | 1.721571 | 1.916108 | 2.181512 | 2.869919 | 2.633652 
23 | 1.576899 | 1.764611 | 1.973587 | 2.206114 | 2.464716 | 2.752166 
24 | 1.608437 | 1.808726 | 2.082794 | 2.288828 | 2.563804 | 2.876014 
25 | 1.640606 | 1.858944 | 2.098778 | 2.368245 | 2.665836 | 3.005434 
26 | 1.673418 | 1.900293 | 2.156591 | 2.445959 | 2.772470 | 3.140679 
27 | 1.706886 | 1.947800 | 2.221289 | 2.581567 | 2.885869 | 3.282010 
28 | 1.741024 | 1.996495 | 2.287928 | 2.620172 | 2.998703 | 3.429699 
29 | 1.775845 | 2.046407 | 2.856566 | 2.711878 | 8.118651 | 3.584036 
30 | 1:811862 | 2.097568 | 2.427262 | 2.806794 | 3.248398 | 3.745318 
31 | 1.847589 | 2.150007 | 2.500080 | 2.905031 | 3.378133 | 3.918857 
02 | 1.884541 | 2.203757 | 2.575083 | 8.006708 | 3.508059 | 4.089981 
33 | 1.922231 | 2.258851 | 2.652835 | 3.111942 | 3.648381 | 4.274030 
34 | 1.960676 | 2.315822 | 2.781905 | 3.220860 | 3.794316 | 4.466362 
35 | 1.999890 | 2.378205 | 2.818862 | 3.883590 | 3.946089 | 4.667348 


Compounp IntTEREsT TABLE. 299 


Showing the amount of $1, or £1; Pano uns interest, from 1 to 35 
years, at the following rates: 


ica] 
a 54 6% 1% 8%. 94 10Z 
gS Dae ARG Tie Mees See eee eR PRENPIPRINI Sols 
1 | 1.050000 | 1.060000 | 1.070000 | 1.080000 | 1.090000 | 1.100000 
2 | 1.025000 | 1.123600 | 1.144900 | 1.166400 | 1.188100 | 1.210000 
3 | 1.157625 | 1.191016 | 1.225043 | 1.259712 | 1.295029 | 1.331000 
4 | 1.215506 | 1.262477 | 1.310796 | 1.360489 | 1.411582 | 1.464100 
5 | 1.276282 | 1.338226 | 1.402552 | 1.469328 | 1.538624 | 1.610510 
6 | 1.840096 | 1.418519 | 1.500730 | 1.586874 | 1.677100 | 1.771561 
7% | 1.407100 | 1.503630 | 1.605782 | 1.718824 | 1.828040 | 1.948717 
8 | 1.477455 | 1.593848 | 1.718186 | 1.850930 | 1.992563 | 2.143589 
9 | 1.551328 | 1.689479 | 1.838459 | 1.999005 | 2.171893 | 2.357948 
10 | 1.628895 | 1.790848 | 1.967151 | 2.158925 | 2.367364 | 2.593743 
11 | 1.710339 | 1.898299 | 2.104852 | 2.331639 | 2.580426 | 2.853117 
12 | 1.795856 | 2.012197 | 2.252192 | 2.518170 | 2.812665 | 3.138429 
13 | 1.885649 | 2.132928 | 2.409845 | 2.719624 | 3.065805 | 3.452271 
14 | 1.979982 | 2.260904 | 2.578534.| 2.937194 | 3.341727 | 3.797499 
15 | 2.078928 | 2.396558 | 2.759032 | 3.172169 | 3.642488 |. 4.177248 
16 | 2.182875 | 2.540852 | 2.952164 | 3 425943 | 3.970806 | 4.594973 
17 | 2.292018 | 2.692773 | 3.158815 | 3.700018 | 4.327633 | 5.054470 
18 | 2.406619 | 2.854389 | 3.379982 | 3.996020 | 4.717120 | 5.559917 
19 | 2.526940 | 3.025600 | 3.616528 | 4.315701 | 5.141661 | 6.115939 
20 | 2.653298 | 3.207186 | 3.869685 | 4.660957 | 5.604411 | 6.727500 
21 | 2.785963 | 3.399564 | 4.140562 | 5.083834 | 6.108808 | 7.400250 
92 | 2.925261 | 3.603537 | 4.430402 | 5.486540 | 6.658600 | 8.140275 
23 | 3.071524 | 3.819750 | 4.740580 | 5.871464 | 7.257875 | 8.954302 
24 | 3.225100 | 4.048935 | 5.072367 | 6.341181 | 7.911083 | 9.849733 
25 | 3.386354 | 4.291871 | 5.427483 | 6.848475 | 8.623081 |10.834706 
26 | 3.555673 | 4.549383 | 5.807353 | 7.396353 | 9.399158 |11.918177 
27 | 3.733456 | 4.822346 | 6.213868 | 7.988062 |10.245082 113.109994 
28 | 3.920129 | 5.111687 | 6.648838 | 8.627106 |11.167140 |14.420994 
29 | 4.116186 | 5.418888 | 7.114257 | 9.317275 |12.172182 |15.863093 
30 | 4.821942 | 5.748491 | 7.612255 |10.062657 |13.267679 |17.449402 
31 | 4.538040 | 6.088101 | 8.145113 |10.867669 |14.461770 |19.194343 
32 | 4.764942 | 6.453387 | 8.715271 |11.737083 |115.763329 |21.113777 
33 | 5.008189 | 6.840590 | 9.325340 |12.676050 |17.182028 |23 225154 
34 | 5.253348 | 7.251025 | 9.978114 |13.690134 |18.728411 |25.547670 


35 | 5.516015 | 7.686087 |10.676582 |14.785344 |20.413968 |28. 102437 


300 CompounD INTEREST. 


Exam. If a boy, at birth, have a legacy of $5000 
left him, how much will he have to receive at the age of 
21, at the rate of 4% compound interest? 


The amount of $1, taken from the table, for 21 years at 4% is 
2.278768, and 2.278768 x 5000 = $11893.84, the required amount. 

Rue Il. Zo find the principal, which, at a. given rate and in a 
given time, would amount to a gwen sum; Or, to find the present 
worth of a sum at compound interest, for a gwen tume and at a given 
rate: Divide the given sum by the amount of $1 for the given time 
and. rate. 


Exam. What sum must be invested at the birth of a 
child, at 4% per annum, compound interest, so as to 
amount to $11893.84 at the end of 21 years? | 

Here, the amount of $1 for 21 years, at 4%, taken from the table 
is 2.278768, and dividing the given sum by this, we have $11393.84 ~ 
2.278768 = $5000. . 

Rute IIL. To find the time in which a given principal will produces . 
a gwen amount. compound interest, at a given rate: Divide the given 
amount by the given principal ; the result is the amount of $1 for the 
time required. IPf this umount be found in the table, the required 
time will be found in the column marked years, opposite said amount. 


Exam. In what time will $5000 amount to $11393.84, 
at 4%, compound interest ? | 
Here, we have $11393.84 + 5000 = 2.278768, the amount of $1 for 


the required time. This number, 2.278768 is found in the table in 
the 4% column, and opposite is found 21 years, the time required. 


If, when the amount is divided by the principal, the result is not 
found in the table, proceed as illustrated in the following 


Exam. In what time will $5000 amount to $10000, 
compound interest, at 6%, in other words, how long will it 
take to have money double itself, at 6%, compound 
interest ? 

Here, dividing $10000 by $5000, we obtain $2, the amount of $1 
for the required time. We now take from the table the nearest 
umount in excess of $2. Looking along the 6% column we find 
2.012197, and looking in the yearly column opposite, we find 12 
years, or the time in which $1 would amount to 2.012197, the dif- 


ference being .012197, the interest which must accrue in the fraction 
of a year on §2. 


Sinking Funp CompvutTATIONS. 301 


Now, the interest on $2 for 1 year at 67 is 12¢, and the question 
now is, if the interest on $2 for 1 year be .12, in what time will $2 
give .012197 interest? For this we have the following proportion: 
As .12: .012197 :: 1 yr. x or the time obtained by dividing .012197 
by .12 = .1016 decimal of a year = 1 mo. 6da., which is deducted from 
12 years as found in the table; the required time is 11 yrs. 10 mo. 
24 da. (12 yrs. — 1 mo. 6 da.). 


A more expeditious method for finding the time when 
a sum of money will donble itself, at any rate per cent., 
compound interest, is as follows: 


RULE. Divide the number 69.3 by the rate per cent. and to the 
result add decimal .35, the result ts the time in years and the decimal 
of a year. 

Thus, taking the foregoing example, to find in what time $5,000 
will double itself : Dividing 69.3 by 6 we get 11.55 to which decimal 
.35 ts added ; the result is 11.90 years. Reducing decimal .90 to 
months and days, we get 11 yrs. 10 mo. 24 days as found above. 


Ruue. To ascertain the sum of money necessary to be laid aside at 
the end of each year, to provide for the payment of a given amount, 
in a given number of years, at a stated rate of interest compounded 
annually: Divide the interest for one year upon the given sum by the 
compound interest of $1 at the given rate, and given number of years; 
the result will be the annual sum. 


Exam. What sum must be put into the sinking fund 
at the end of each year at 4 per cent. compound interest, 
to provide for the payment of $20000, payable in 10 
years? 


Here, the interest of $20000 for one year at 4% = $800; and the 
amount of $1 for ten years at 4% compound interest (taken from 
comp. int. table, page 298) is 1.480244. Deducting $1 from this 
amount, the result is the compound interest, .480244. 

Applying the rule, now, we have: $800 + .480244 = $1665.82, 
the required sum. 

To prove that the work is correct, add to $1665.82 its interest for 
a year at 47, and to the result add the sum to be laid aside ($1665.82) 
make the sum a new principal, to which add its interest for a year 
at 4%, and to the sum add $1665.82 for the next principal. Repeat 
the process until you have found $20000. The work is left for 
the student. 

If the money is to be laid aside at the beginning of each year; 
divide the result found above by the amount of $1 for one year at 
the given rate; thus, $1665.82 + 1.04 = $1601.75, the sum required 
to be laid aside at the beginning of the yeer. 


BOND COMPUTATIONS. 


RuLE. Zo find the present value of Bonds: (1) Divide $1 by tts 
amount at compound interest for the given time and at the Cesired rate. 
(2) Subtract the result thus found from $1, and multiply the remainder 
by the difference of rates. (3) Dwide this last result by the desired 
rate; this will give either a premium or a discount. 

If the desired rate be greater than the bond rate, tt will be a discount 
to be subtracted from $1; and if less, it will be a premium to be added; 
the result, in either case, will be the ‘value of $1, and 100 tumes this the 
value of the bond. 


Exam. 1. What must be paid fora bond maturing in 
10 years with 4% interest,.payable annually, to pay the 
purchaser 6%. 


The amount of $1 @ 6% compound interest for 10 years = 1. 790848. 
(Taken from comp. interest table, page 298.) ‘ 
Now. 1 + 1.790848 = .558895; next, 1 —.558395 (1.000000 558395) 
= .441605 then, .441605 x 2 (6¢—47) = .883210; and .8838210 + 6= 
.147202, which is a discount to be subtracted from $1, thus, 
1 — .147202 = .852798; this last is the value of $1 on the proposed 

terms; and 100 times that = $85.2798, the value of the bond. 


Exam. 2. If the interest were paid semi-annually, 
instead of annually, in the foregoing, what would be 
the value of the bond ? 


Referring to the compound interest eae page 298, we find the 
amount of $1 for 20 years @ 34 to be 1.806111. 

Now, 1 + 1.806111 = .553676; and 1— .558676 = .446324; next, 
.446324 x 2 = .892648 then, .892648 + 6 = .148775, which is to be 
subtracted from $1, thus, ea 148775 = 851225, and 100 times this © 
= $85. 1225. 


Exam. 8. What must be paid for bonds amounting to 
$10000 maturing in 10 years, with interest @ 6%, to net 
4%, interest payable semi-annually ? 


Here, the amount of $1 for 20 years @ 2¢ (taken from the com 
int. table, page 298) is found to be 1.485947. - 

Now, 1 + 1.485947 = .672971; next, 1 — .672971 = .827029; and 
327029 X 2 = .654058. Dividing this by the net rate, we have. 
.654058 + 4 = .163514, which is a premium to be added to $1, thus, 
1 + .163514 = 1.163514, the value of $1 at the proposed terms; and 
10000 times 1.163514 = $11635.14, the value of the bonds. 


Bonp CompvuTATIONS. 303 


Exam. 4. What must be paid for a 5% bond maturing 
in 4 years, to net 4%, interest payable annually? 


In this, the amount of $1 taken from the int. table for 4 years 
@ 4% = 1.169859. And 1 + 1.169859 = .854804: now, 1 — .854804 = 
.145196 (the difference of rates being 1, there is no need to multiply) 
so we divide this by the required net rate, thus, .145196 + 4 =.036299 
which is a premium to be added giving 1.036299, the value of $1. 
and 100 times that = $103.6299 or $103.63 the cost of the bond. 


To prove this to be correct, add to $103.63 its interest for a year at 4% = $4.14» 
the amount is $107.77, from*which deduct $5 (bond int.) the result is $102.77: 
Compute a year’s interest on this principal and deduct the bond interest; repeat 
the process until at maturity the value is found to be $100. 


ANNUITIES. 


RuLE. To find the amount of an annuity, payable yearly, the pay- 
ments of which are forborne for a given time, compound interest being 
charged on them as they became due, 


(1) Divide the compcund interest of $1 for the given time and rate, 
by the rate per cent.; the result will be the amount of an annuity of $1 
forborne for the proposed time. 


(2) Multipiy this annuity by the given annuity; the result will be the 
amount required, 


Exam. If a person save $500 a year, and improve it 
at 4% per annum, compound interest; how much will he 
be worth at the end of 21 years? 


In this, the amount of $1 for 21 years, at 44 compound interest is 
2.2;87683. And deducting $1 from this we have the compound 
interest of $1 = 1.278768. : 


Applying the rule, we have 1.278768 + .04 x 500 = $15984.60, 
what the person is worth at the end of 21 years.. 


Proor: What sum must be laid aside at the end of 
each year, for 21 years, at 4% compound interest, to 
amouut to $15984.60 ? 


Applying the rule for the sinking fund, given on page 301, we find 
the interest of $15984.60 for 1 year at 4% = $639.38, and dividing this 
by the compound interest of $1 we have: $689.38 + 1.278768 = $500. 


To Discuarazt A Given Dest in Severat EQuar 
Payments, Inctupina PrincipAL AND INTEREST, AT 
Any Rare Per CEnt. 


Ruue. (1) Multiply the amount of the debt at compound interest 
for the given time and rate, by the rate per cent. (2) Divide the result 
by the compound interest of $1 for the given time and rate; the result 
will be the payment. 


Exam. A manufacturer sold 5 pianos at $720 each, 
agreeing to take in settlement six equal half yearly pay- 
ments, including principal and interest, at 6% per annum; 
what was the amount of each payment! ? 


Solution.— $720 xX 5 = $3600, the total cost. 


The yearly rate being 6%, the rate half yearly is 3%, and there are 
to be 6 equal payments. Now, the amount of $1 for the first interval 
is $1.03, which is to be raised to the 6th power; thus, 1.03° = 1.03 x 
1.08 « 1.03 & 1.03 K 1.08 * 1.03 = 1.194052, which is the amount 
of $1 at compound interest for the given time and rate; and deduct- 
ing $1 from this amount, gives the compound interest .194052, on $1 
for the given time and rate. 


1.194052 x 8600 X.08 _ gee4 55, 


Applying the rule, now, ha 
pplying w, we have: “T54058 


the amount of each payment. 


Notrs.— The contracted method for the multiplication of decimals given on 
. pages 293 and 294 can be used in raising 1.03 to the 6th power. Or, the amount 
of $1 for 6 years at 3% may be had at a glance from the compound interest 
table, pages 298 and 299. 

Such problems as the foregoing, however, can be more easily and 
rapidly solved by the use of the table given on pages 306 and 307, 
where the amount required to discharge the debt of $1, in equa: 
payments, is given. 


To illustrate, let us take the foregoing example. Referring to the 
table, we look for 6 years, and opposite, under 327, is the decimal 
.184598. This is the amount required to discharge $1, in equal 
payments, in 6 years, at 8%: and multiplying this by the given debt, 
we obtain the amount of each equal aie thus, .184598 x 3600 = 
$664.55, as found by the male 


To Discoarce a Given Dest, Erc., Ero. 3805 


In connection with the foregoing rule and example 
we subjoin the following: | 


PROOF. 

$3600. Brought forward $1879.75 

Int. on $3600 for 6 mo. at83% = 108 Int. on $1879.75; 6 mo. 3% = 56.39 
$3708 $1936.14 

Amt, of Ist payment = 664.55} Amt. of 4th payment — 664.55 
$3043.45 $1271.59 

Int. on $3043.45; 6 mo, 3% = 91.30 | Int. on $1271.59; 6 mo. 3% = 38.15 
$3134.75 $1309.74 

Amt. of 2d payment — 664.55 | Amt. of 5th payment — 664 55 
$2470.20 poto. 19 

Int. on $2470.20; 6mo. 3% = 74.10 | Int. on $645.19; 6 mo. 8% = 19.36 
$2544.30 $664 . 55 

Amt, of 83d payment — 664.55 | Amt. of 6th or last payment — 664.55 

Carried forward $1879.75 
As a further illustration of the rule, we give the fol- 


lowing: 

Exam. To settle a debt of $6,000, three notes, each for 
an ‘equal sum, and drawn at 4 months, are given; what 
is the amount of each note, including principal and 
interest, at 6% per annum ? 

Solution. — The annual rate being 6%, the rate for 4 mo. is 2¢ , 
snd there are to be 3 equal payments. 

{he amount of $1 for the first interval of 4 mo. at 2% (.02) is 1.02; 
and this raised to the 3d. power gives the amount of $1 for the pro- 


posed time and rate; thus, 1.02 x 1.02 X 1.02 = 1.061208; and de- 
ducting $1 from this amount gives the compound interest of $1, viz. 


.061208. 
Applying the rule, now we have: ee = $2,080 .528 


the amount of each note. 

Now, by referring to the table on the next page, and looking in the 
yearly column for 8 years (3 payments) and opposite, in the column 
for 2% we find decimal .346755; multiplying this by the debt, we get 
the amount of each note, including principal and interest; thus 
.846755 X 6,000 = $2,080.53. (The proof is left for the student.) 


Solve the following problem by using the table: 


A debt of $50,000 is to be paid in 4 years, in 8 equal semi-annual 
payments; what is the amount of each payment, including principal 
and interest, at 5% per annum ? 

Turning to the table, we ook for 8 years (8 payments) and opposite, 
in the 24% column (half of 5%), we find decimal .139467, and multi- 
plying this by the debt, we have .139467 x 50,000 = $6,973.35, the 
required amount. 


306 Dest TABLE. 


. 


* 


Table showing the amount required to discharge a debt of $1 in 
equal payments, from 2 to 21 years, at the rates given. , 


——_— 


——— | — | | fe | | 


2 | .515049 | .518827 | .522611 | .526400 | .530196 2 
8 | .846755 | .850185 | .3853530 | .356930 | .860848 8 
4 | .262625 | .265818 | .269022 | .272250 | .275494 & 
5 | .212160 | .215247 | .218855 | .221482 | .224627 5 
6 6 
7 7 
8 8 


.178526 | .181550 | .184598 | .187668 | .190762 
. 154512 | .157495 | .160506 | .163544 | .166609 
.136510 | .189467 | .142456 | .145476 | .148528 
9 | .122515 ! .125453 | .128106 | .181446 | .184493 9 
10 | .111827 | .114258 | .117231 | .120241 | .128291 10 


11 .102178 | .105106 | .108077 | .111092 | .114149 11 
12 | .094559 | .974871 | .100462 | .103484 | .106552 12 
13 | .088118 | .910483 | .094030 | .097062 | .100142 18 
14 | .082602 | .860208 | .088526 | .091571 | .094669 14 
15 | .077826 | .807656 | .088767 | .086825 | .088941 15 


16 | .078650 | .765989 | .079611 | .082685 | .085820 16 
17 | .069970 | .729278 | .075953 | .079048 | .082198 | 17 
18 | .066702 | .696701 | .072709 | .075871 | .078993 18 
19 | .068782 | .667606 | .069814 | .072940 | .0761388 19 
20 | .061157 | .641471 | .067216 | .070361 | 078582 20 
21 .058785 | .617872 | .064872 | .068037 | .071280 21 


As an aid to the solution of problems of the foregoing nature, 
recourse can be had to the Compound Interest Tables given on pages 
298 and 299. 


Exam. Suppose it were required to pay off a debt of 
$40000 in two years, in 8 equal quarterly payments, in- 
cluding principal and interest, at 8% per annum; what 
should be the amount of each payment? 


Solution.— The yearly rate being 87%, the quarterly rate is 2 (.02) 
and there are to be 8 payments. The amount of $1 for the first 
interval is $1 plus .02 = 1.02; and for 8 intervals, the amount of $1 
is 1.02°; 1.02 raised to the 8th power, requiring seven multiplications 
to get the amount. To avoid this labor, we refer to the interest 
table, and looking in the yearly column for 8 years (8 payments) and 
opposite, in the 2% rate column, we find 1.171659, the amount of $1 


Dest TABLE. 307 


Table showing the amount required to discharge a debt of $1 in 
equal payments, from 2 to 21 years, at the rates given. 


Years 4174 5% 64. 1% 8% Years 
2 .53838998 | .587805 | .545487 | .553091 | .560769 2 
3 .863774 | .3867208 | .874109 | .881051 | .888034 3 
4 .2787438 | .282012 | .288591 | .295228 | .3801921 4 
5 227792 | .230974 | .287396 | .248890 | .250457 5 
6 .198878 | .197017 | .208863 | .209796 | .216315 6 
7 .169701 | .172819 | .179135 | .185553 | .192072 7 
8 .151610 | .154722 | .1610386 | .167607 | .174015 8 


9 | .187572 | .1406990 | .147022 | .158486 | .160079 9 
10 | .126821 | .129506 ; 185881 | .147547 | .149029 10 


11 .117248 | .120889 | .126793 | .1383357 | .140076 11 
12 | .109666 | .112825 | .119278 | .125902 | .182695 12 
13 | .108275 | .106456 | .112960 | .119651 | .126522 13 
14 | .097820 | .101024 | .107759 | .114845 | .121297 14 
15 | .098114 | .096343 |-.102963 | .109795 | .112226 15 


16 | .089015 | ,092269 | .098952 | .105857 | .112098 16 
17 | .085418 | .088699 | .095445 | .102425 | .109629 17 
18 | .082237 | .085546 | .092856 | .099413 | .196702 18 
19 | ,079407 | .082090 | .089621 | .096753 | .104128 19 
20 | .U76876 | .080243 | .087185 | .0943893 | .101852 20 
21 .074601 | .077996 | .085005 | .092286 | .0998382 21 


for the proposed time and rate; and deducting $1 from this amount, 
we get .171659, the compound interest. 
1.171659 x 40000 X .02 _ 


N by th ] 304, h 
ow, by the rule on page we have erent = 


$5460.42, amount of each payment. 


The problem can be more readily solved, however, by using the 
Debt Tabie, as follows: | 

Looking in the yearly column for eight years (8 payments) and 
opposite in the 2% rate column, we find decimal .136510, the amount 
required to discharge the debt of $1 in the proposed time at the 
given rate; and this multiplied by the debt gives the payment, at 
once; thus, .186510 « 40000 = $5460.40. 


Norr. — There is a discrepancy of 2¢ in the answer given above; but if the 
decimal in the interest table were carried out to eight places, 171165938, 
instead of six, the answer would be $5460.39 + a little in excess of 39¢. Six 
places of decimals, however, are sufficient for all practical purposes. 


308 A SrmveteE MeEtHop FoR THE EXTRACTION OF 


THE CUBE ROOT. 


Rue I.— Point off the given number into periods of three figures 
each, beginning at the units. 

Il. Find the greatest cube of the left-hand period, and write its root 
as the first figure of the required root. 

Ill. On the left of the dividend write three times the square of the 
first figure of the root, and annex two ciphers, or dots to represent 
them; this number is trial divisor by which find the neat figure of the 
root. 

IV. On the left of the trial divisor, and one line lower, write the root 
figure thus found in the place of units, and three times the part of the 
root formerly found write in the place of tens, hundreds, ete. 

V. Multiply this last number by the last figure of the root, set the 
result under the trial divisor, and add the two for the complete divisor. 

VI. Multiply this complete divisor by the quotient figure, subtract 
and bring down another period. . 

VIL. Lastly, underneath the complete divisor, write the square of 
the last figure found in the root; add this square and the two numbers 
immediately above tt; annex two ciphers (dots) and the result will be 
another trial divisor, with which proceed as before 


Nota.— If a cipher occur in the root, annex two moro ciphers (dots) to the trial 
divisor, and another period to the dividend; then proceed as before. 


Exam. Extract the cube root of 389816897625. 


889816897625( 7305 
y 49 348 
3 46816 

147°: +. d. 46017 
213 639 “799897625. 

15339 c. dd. 799897625 

9 
15987°-*- +t. d. 
109525 

21905 ~ 159979525 ec. d. 


Having proceeded as far as the second figure (8) of the root, we 
add its square (9) to the ¢wo numbers (639 and 15339) immediately 
above it, getting 15987 to which we annex two ciphers, or dots to 
represent them, and we have 1589700 (15897: -) for trial divisor (t. d.). 
We next bring down another period (897) to the dividend and we 
have 799897. The trial divisor is not contained in this number, so 
we set a cipher in the root, annex two more ciphers (dots) to the 
et divisor, and another period (625) is brought down to the 

ividend, . 


« 


Tue Cuse Roor. 309 


The trial divisor (t. d.) is now 158970000 (15897°*--) by which we 
find the next figure (5) in the root. Setting 5, now, to the left of 
the trial divisor, and one line lower, in the place of units, and 
5 times 730, the part of the root formerly found, in the place of tens, 
hundreds, etc., we get 21905. 

Finally, 5 times 21905, or 109525 is added to the trial divisor 
which gives 159979525 for the complete divisor (c, d.); and 7805 is 
the required root, 


Exam. Required the cube root of 477 to four places 
of decimals. 


7 49 477 (7.8183 
3 348 
~ 147°° t. d. 134000 
218 1744 131552 
16444 c. d. 2448000 
64. 1827541 
18252": t.d. 620459000 
2341 2341 549175797 
1827541 + od 71283203000 
1 54940781637 
1829883°- t. d. 
93433 70299 ) 
[505850 ce. d, 
9 
183128907°- t. d. 
234393 703179 
~ 18313593879 


Having found 7, the first figure of the root, 3 times its square, 147, 
with two ciphers (dots) annexed, is trial divisor. Annexing three 
ciphers (a period) to the dividend,we find the next figure, 8, of the 
root, inserting the decimal point. Then 8 is set to the left of 
the trial divisor, and one line lower, in the place of units, and three 
times 7, the first root figure, or 21, is set in the place of tens and 
hundreds, giving 218. This is now multiplied by 8, and the result, 
1744, is added to the trial divisor, giving 16444 for the complete 
divisor. 8 times this is now set under the dividend, and the result, 
131552, subtracted. Finally, the square of 8 = 64, is set under the 
complete divisor and added to the two numbers immediately above 
it (included in the brace), and the sum of the three numbers with 
two ciphers (dots) annexed, is the next trial divisor with which find 
the next figure of the root, and proceed as before. 


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